^'i 


IN  MEMORIAM 
FLORIAN  CAJORI 


"if^l^yU^   da^p-^, 


ELEMENTS 


OF 


TEIGONOMETKY 


WITH    TABLES 


BY 


HERBERT  C.  WHITAKER,  Ph.D. 

CENTRAL   MANUAIi  TRAINING   SCHOOL 
PHILADKLPHLA.,   PENNSYLVANIA 


PHILADELPHIA 
D.    ANSON    PARTRIDGE 

1898 


Copyright,  1898, 
By  H.   C.   WHITAKEK. 


Nortoooti  ?3w8S 

J.  S.  Gushing  &  Co.  —  Berwick  Sc  Smith 
Norwood  Mass.  U.S.A. 


PREFACE. 


The  introduction  and  first  five  chapters  of  this  book  have 
been  prepared  for  the  use  of  beginners  in  the  subject.  The 
second  appendix  has  been  added  for  those  intending  to  take 
up  work  in  higher  departments  of  Mathematics. 

The  subject  of  logarithms  is  usually  considered  in  treatises 
on  algebra ;  but  for  the  convenience  of  those  not  familiar 
with  the  theory  of  this  subject,  the  first  appendix  has  been 
added. 

To  aid  in  a  clear  understanding  of  each  process  in  the 
various  investigations,  the  aim  has  been  to  closely  associate 
with  every  equation  a  definite  meaning  with  reference  to  a 
diagram. 

The  tables  used  are  in  general  as  precise  as  the  data 
usually  obtained  by  students  in  engineering,  physics,  or 
chemistry.  It  is  assumed  that  a  few  seven-place  tables  are 
accessible  to  a  class ;  and  it  is  recommended  that  some  class- 
room work  be  done  with  seven-place  tables,  especially  in 
the  solution  of  spherical  triangles. 

It  is  hoped  that  the  answers  are  free  from  error.  The 
numerical  computations  were  made  with  care  by  the  author, 
but  the  results  have-not  been  verified  by  any  one  else. 

Any  corrections  or  suggestions  relating  to  the  book  will 
be  thankfully  received. 

H.  C.  WHITAKER. 

Philadelphia,  May,  1898. 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsoftrigonOOwhitrich 


CONTENTS. 


Table  of  natural  values  of  functions i 

Table  of  circular  measure  of  angles vii 

Table  of  hyperbolic  functions vii 

INTRODUCTION. 

§  1.     Meaning  of  geometric  operations 1 

§  2.     Negative  segments  and  angles 3 

§  3.     Products  and  quotients  of  segments 5 

§  4.     Homogeneity  of  trigonometric  equations  ....  6 

§  5.     Significant  figures  of  numbers 7 

§  6.     Approximate  values  of  numbers 7 

§  7.     Contractions  in  arithmetical  operations     ....  7 


CHAPTER  I. 

Functions  of  Angles. 

§  8.     Generation  of  an  angle 11 

§  9.     The  four  quadrants 12 

§  10.     The  six  ratios 14 

§  11.     Relations  between  functions  and  segments        ...  17 

§  12.     Device  for  memorizing  relations 19 

§  13.     Absolute  values  and  algebraic  signs  of  functions       .        .  20 

§  14.     Functions  of  180°  ±  x,  of  90°  +  x,  and  of  -  x  .        .        .  22 

§  15.     The  curve  of  each  function 24 

§  16.     Limiting  values  of  functions 26 

§  17.     The  number  of  angles  corresponding  to  a  value  of  a 

function 26 

§  18.     Functions  oi\x  and  2  x 28 

§  19.     Functions  of  particular  angles 29 


CONTENTS. 

CHAPTER  II. 
Explanation  of  the  Tables. 

PAGK 

§  20.  The  table  of  natural  values  of  functions    ....  31 

§  21.  Principles  of  logarithms 37 

§  22.  Use  of  the  table  of  logarithms 41 

§  23.  The  table  of  logarithms  of  functions 45 

CHAPTER   III. 
Right  Triangles. 


§  24.  Problems  are  indeterminate,  ambiguous,  determinate,  or 

impossible 

§  25.  Relations  of  the  parts  of  a  right  triangle 

§  26.  Solutions  of  triangles  using  numbers. 

§  27.  Applications 

§  28.  Area  of  a  right  triangle 

§  29.  Solutions  of  triangles  using  logarithms 

S  30.  Problems 


46 
47 
49 
51 
55 
56 
58 


CHAPTER  IV. 

The  Isosceles  Triangle  and  the  Regular  Polygon. 

§  31.  Formulse  for  isosceles  triangles  .         ...         .         .         .60 

§  32.  Problems  in  isosceles  triangles 61 

§  33.  Formulse  for  regular  polygons 63 

§  34.  Problems  in  regular  polygons 65 

CHAPTER  V. 

Oblique  Triangles. 

§  35.     Notation 66 

§  36.     Methods  of  solution 66 

§  37.     Solutions  using  numbers 70 

§.38.     Cot .5  =  0^^^^  -cot J. 74 

^  b 


CONTENTS. 


PAGK 

§  39.  The  area  of  a  triangle 76 

§  40.  Solutions  using  logarithms 77 

§  41.  Applications  with  solutions 81 

§  42.  Problems 87 


CHAPTER  VI. 
Pboperties  of  Triangles. 

§  43.     The  altitudes,  bisectors,  and  medians        ....  94 

§  44.    The  incircle  and  the  excircles 96 

§  45.     The  circumcircle 98 

§46.    The  sum  and  difference  of  two  sides  and  two  angles .        .  98 

§  47.     General  method  of  solution,  given  three  segments     .        .  100 
§  48.     Points  of  intersection  of  the  circumcircle  .        .        .        .100 

§  49.     The  nine-points  circle 102 

§  50.     Abscissae  and  ordinates  of  points 104 

§  51.     Problems 105 

§  62.     Relations  between  the  angles  of  a  triangle         .        .        .  108 


CHAPTER  VII. 
Spherical  Triangles. 

53.     Tlie  general  spherical  triangle 110 

54!     Formulae  for  right  triangles 112 

55.  Device  for  rapidly  obtaining  formulae        .         .        .        .114 

56.  Solution  of  right  triangles 115 

57.  Polar,  quadrantal,  and  isosceles  triangles ;    the  regular 

polygon 117 

58.  Oblique  spherical  triangles 120 

59.  The  incircle  and  the  excircles 120 

60.  The  incircle  of  the  polar  triangle  and  the  circumcircle      .  121 

61.  Formulae  for  oblique  triangles 122 

62.  Methods  of  solution 122 

63.  Solutions    .         . 123 

64.  A  desired  part  directly  in  terms  of  given  parts          .         .  126 

65.  Applications  to  geodesy  and  astronomy     .        .        .        .128 

66.  Area  of  a  spherical  triangle 132 


CONTENTS. 

APPENDIX  I. 

Theory  of  Logarithms. 

§  67.  Arithmetic  and  geometric  series 

§  68.  Napierian  logarithms  treated  geometrically 

§  69.  Present  systems  treated  geometrically 

§  70.  Natural  logarithms 

§  71.  The  number  corresponding  to  a  given  logarithm 

§  72.  A  logarithm  is  an  exponent  of  a  base 

§  73.  The  logarithm  corresponding  to  a  given  number 

§  74.  The  logarithmic  curve 

S  75.  Problems 


133 
134 
136 
137 
138 
140 
141 
143 
144 


APPENDIX  II. 

GONIOMETRY,    COMPLEX    QUANTITIES,    HYPERBOLIC    FUNCTIONS. 

§  76.     Functions  of  the  sum  or  difference  of  angles     .        .        .  144 

§  77.     Trigonometric  equations  ;  the  quadratic  equation     .         .  146 

§  78.     The  circular  measure  of  an  angle 149 

§  79.     Arc  functions 150 

§  80.     Complex  Quantities  :  addition,  subtraction,  multiplication, 

division,  powers  and  roots 152 

§  81.     Functions  of  multiple  angles 159 

§  82.     Exponential  values  of  sin  6  and  cos  d         .        .         .        .  159 

§  83.     Hyperbolic  Functions :  the  equilateral  hyperbola     .        .  160 

§  84.     Definitions  of  hyperbolic  functions 161 

§  85.     Exponential  values  of  hyperbolic  functions        .         .         .  162 

§  86.     General  relations 163 

§  87.     Longitude  of  w 164 

§  88.     Solution  of  the  cubic  equation 164 

Problems  for  Examination 167 

Answers  to  Problems 177 

Table  of  Logarithms  of  Numbers viii 

Table  of  Logarithms  of  Functions x 

Table  of  Formula xvi 

Table  of  Constants xvi 


A   TABLE 

OF   THE 

NATURAL  VALUES  OF  THE  TRIGONOMETRIC 
FUNCTIONS  OF  ANGLES 

FROM   0°  TO  90°  FOR  EVERY  10' 


A   TABLE 

FOR  CONVERTING  THE  SEXAGESIMAL  MEASURE 

OF  AN  ANGLE  INTO  CIRCULAR  MEASURE 


A  TABLE 

FOR  CONVERTING  CIRCULAR  FUNCTIONS  INTO 

HYPERBOLIC   FUNCTIONS 


NATURAL  VALUES  OF  FUNCTIONS. 


X 

sinjif 

tanjf 

secAf 

CSCJf 

cotjr 

cosjr 

0° 

.0  00000 

0  00000 

1.00  00 

Infinite.  Infinite. 

L.OOOO 

90° 

10' 

02909 

02909 

00 

343.775  343.774 

00 

50' 

20' 

05818 

05818 

00 

171.888 

171.885 

00 

40' 

30' 

08727 

08727 

00 

114.593 

114.589 

00 

30' 

40' 

11635 

11636 

01 

85.946 

85.940 

.99  99 

20' 

50' 

14544 

14545 

01 

68.757 

68.750 

99 

10' 

1° 

.0  1745 

.0  1746 

1.00  02 

57.299 

57.290 

.99  98 

89° 

10' 

2036 

2036 

02 

49.114 

49.104 

98 

50' 

20' 

2327 

2328 

03 

42.976 

42.964 

97 

40' 

30' 

2618 

2619 

03 

38.202 

38.188 

97 

30' 

40' 

2908 

2910 

04 

34.382 

34.368 

96 

20' 

50' 

3199 

3201 

05 

31.258 

31.242 

95 

10' 

2° 

.0  3490 

.0  3492 

1.0006 

28.654 

28.636 

.99  94 

88° 

10' 

3781 

3783 

07 

26.451 

26.432 

93 

50' 

20' 

4071 

4075 

08 

24.562 

24.542 

92 

40' 

30' 

4362 

4366 

10 

22.926 

22.904 

90 

30' 

40' 

4653 

4658 

11 

21.494 

21.470 

89 

20' 

50' 

4943 

4949 

12 

20.230 

20.206 

88 

10' 

3° 

.0  5234 

.0  5241 

1.00  14 

19.107 

19.081 

.99  86 

87° 

10' 

5524 

5533 

15 

18.103 

18.075 

85 

50' 

20' 

5814 

5824 

17 

17.198 

17.169 

83 

40' 

30' 

6105 

6116 

19 

16.380 

16.350 

81 

30' 

40' 

6395 

6408 

21 

15.637 

15.605 

80 

20' 

50' 

6685 

6700 

22 

14.958 

14.924 

78 

10' 

40 

.0  6976 

.0  6993 

1.00  24 

14.336 

14.301 

.99  76 

86° 

10' 

7266 

7285 

26 

13.763 

13.727 

74 

50' 

20' 

7556 

7578 

29 

13.235 

13.197 

71 

40' 

30' 

7846 

7870 

31 

12.745 

12.706 

69 

30' 

40' 

8136 

8163 

33 

12.291 

12.250 

67 

20' 

50' 

8426 

8456 

36 

11.868 

11.826 

64 

10' 

5° 

.0  8716 

.0  8749 

1.00  38 

11.474 

11.430 

.99  62 

85° 

10' 

9005 

9042 

41 

11.105 

11.059 

59 

50' 

20' 

9295 

9335 

43 

10.758 

10.712 

57 

40' 

30' 

9585 

9629 

46 

10.433 

10.385 

54 

30' 

40' 

9874 

9923 

49 

10.128 

10.078 

51 

20' 

50' 

.1  0164 

.10216 

52 

9.8391 

9.7882 

48 

10' 

6° 

.1  0453 

.10510 

1.00  55 

9.5668 

9.5144 

.99  45 

84° 

10' 

0742 

0805 

58 

9.3092 

9.2553 

42 

50' 

20' 

1031 

1099 

61 

9.0652 

9.0098 

39 

40' 

30' 

1320 

1394 

65 

8.8337 

8.7769 

36 

30' 

40' 

1609 

1688 

68 

8.6138 

8.5555 

32 

20' 

50' 

1898 

1983 

72 

8.4047 

8.3450 

29 

10' 

7° 

.1  2187 

.1  2278 

1.00  75 

8.2055 

8.1443 

.99  25 

83° 

cos/ 

cot/ 

cscy 

secy 

tan/ 

sin/ 

/ 

NATURAL  VALUES  OF  FUNCTIONS. 


X 

sinjif 

tanjr 

secjf 

csc;r 

cotr 

COSJf 

70 

.1  219 

.1228 

1.00  75 

8.  2055 

8.  1443 

.99  25 

83° 

10' 

248 

257 

79 

0156 

7.  9530 

22 

50' 

20' 

276 

287 

82 

7.  8344 

7704 

18 

40' 

30' 

305 

317 

86 

6613 

5958 

14 

30' 

40' 

334 

346 

90 

4957 

4287 

11 

20' 

50' 

363 

376 

94 

3372 

2687 

07 

10' 

8° 

.1392 

.1405 

1.00  98 

7.  1853 

7.  1154 

.99  03 

82° 

10' 

421 

435 

1.01  02 

0396 

6.  9682 

.98  99 

50' 

20' 

449 

465 

07 

6.  8998 

8269 

94 

40' 

30' 

478 

495 

11 

7655 

6912 

90 

30' 

40' 

507 

524 

15 

6363 

5606 

86 

20' 

50' 

536 

554 

20 

5121 

4348 

81 

10' 

9° 

.1564 

.1584 

1.01  25 

6.  3925 

6.  3138 

.98  77 

81° 

10' 

593 

614 

29 

2772 

1970 

72 

50' 

20' 

622 

644 

34 

1661 

0844 

68 

40' 

30' 

650 

673 

39 

0589 

5.  9758 

63 

30' 

40' 

679 

703 

44 

5.  9554 

8708 

58 

20' 

50' 

708 

733 

49 

8554 

7694 

53 

10' 

10° 

.1736 

.1763 

1.01  54 

5.  7588 

5.  6713 

.98  48 

80° 

10' 

765 

793 

60 

6653 

5764 

43 

50' 

20' 

794 

823 

65 

5749 

4845 

38 

40' 

30' 

822 

853 

70 

4874 

3955 

33 

30' 

40' 

851 

883 

76 

4026 

3093 

27 

20' 

50^ 

880 

914 

81 

3205 

2257 

22 

10' 

11° 

.1908 

.1944 

1.01  87 

5.  2408 

5.  1446 

.98  16 

79° 

10' 

937 

974 

93 

1636 

0658 

11 

50' 

20' 

965 

.2  004 

99 

0886 

4.  9894 

05 

40' 

30' 

994 

035 

1.02  05 

0159 

9152 

.97  99 

30' 

40' 

.2  022 

065 

11 

4.  9452 

8430 

93 

20' 

50' 

051 

095 

17 

8765 

7729 

87 

10' 

12° 

.2  079 

.2  126 

1.02  23 

4.  8097 

4.  7046 

.97  81 

78° 

10' 

108 

156 

30 

7448 

6382 

75 

50' 

20' 

136 

186 

36 

6817 

5736 

69 

40' 

30' 

164 

217 

43 

6202 

5107 

63 

30' 

40' 

193 

247 

49 

5604 

4494 

57 

20' 

50' 

221 

278 

56 

5022 

3897 

50 

10' 

13° 

.2  250 

.2  309 

1.02  63 

4.  4454 

4.  3315 

.97  44 

77° 

10' 

278 

339 

70 

3901 

2747 

37 

50' 

20' 

306 

370 

77 

3362 

2193 

30 

40' 

30' 

334 

401 

84 

2837 

1653 

24 

30' 

40' 

363 

432 

91 

2324 

1126 

17 

20' 

50' 

391 

462 

99 

1824 

0611 

10 

10' 

14° 

.2  419 

.2  493 

1.03  06 

4.  1336 

4.  0108 

.97  03 

76° 

cosy 

cot/ 

esc/ 

sec/ 

tan/ 

sin/ 

/ 

Ill 


NATURAL  VALUES  OF  FUNCTIONS. 


X 

sinx 

tanx 

secjr 

cscx 

cotjr 

COSJf 

14° 

.2  419 

.2  493 

1.03  06 

4.  1336 

4.  0108 

.97  03 

76° 

10' 

447 

524 

14 

0859 

3.  9617 

.96  96 

50' 

20' 

476 

555 

21 

0394 

9136 

89 

40' 

30' 

504 

586 

29 

3.  9939 

8667 

81 

30' 

40' 

532 

617 

37 

9495 

8208 

74 

20' 

50' 

560 

648 

45 

9061 

7760 

67 

10' 

15° 

.2  588 

.2  679 

1.03  53 

3.  8637 

3.  7321 

.96  59 

75° 

10' 

616 

711 

61 

8222 

6891 

52 

50' 

20' 

644 

742 

69 

7817 

6470 

44 

40' 

30' 

672 

773 

77 

7420 

6059 

36 

30' 

40' 

700 

805 

86 

7032 

5656 

28 

20' 

50' 

728 

836 

94 

6652 

5261 

21 

10' 

16° 

.2  756 

.2  867 

1.04  03 

3.  6280 

3.  4874 

.96  13 

74° 

10' 

784 

899 

12 

5915 

4495 

05 

50' 

20' 

812 

931 

21 

5559 

4124 

.95  96 

40' 

30' 

840 

962 

29 

5209 

3759 

88 

30' 

40' 

868 

994 

39 

4867 

3402 

80 

20' 

50' 

896 

.3  026 

48 

4532 

3052 

72 

10' 

17° 

.2  924 

.3  057 

1.04  57 

3.  4203 

3.  2709 

.95  63 

73° 

10' 

952 

089 

66 

3881 

2371 

55 

50' 

20' 

979 

121 

76 

3565 

2041 

46 

40' 

30' 

.3  007 

153 

85 

3255 

1716 

37 

30' 

40' 

035 

185 

95 

2951 

1397 

28 

20' 

50' 

062 

217 

1.05  05 

2653 

1084 

20 

10' 

18° 

.3  090 

.3  249 

1.05  15 

3.  2361 

3.  0777 

.95  11 

72° 

10' 

118 

281 

25 

2074 

0475 

02 

50' 

20' 

145 

314 

35 

1792 

0178 

.94  92 

40' 

30' 

173 

346 

45 

1515 

2.  9887 

83 

30' 

40' 

201 

378 

55 

1244 

9600 

74 

20' 

50' 

228 

411 

66 

0977 

9319 

65 

10' 

19° 

.3  256 

.3  443 

1.05  76 

3.  0716 

2.  9042 

.94  55 

71° 

10' 

283 

476 

87 

0458 

8770 

46 

50' 

20' 

311 

508 

98 

0206 

8502 

36 

40' 

30' 

338 

541 

1.06  08 

2.  9957 

8239 

26 

30' 

40' 

365 

574 

19 

9713 

7980 

17 

20' 

50' 

393 

607 

31 

9474 

7725 

07 

10' 

20° 

.3  420 

.3  640 

1.06  42 

2.  9238 

2.  7475 

.93  97 

70° 

10' 

448 

673 

53 

9006 

7228 

87 

50' 

20' 

475 

706 

65 

8779 

6985 

77 

40' 

30' 

502 

739 

76 

8555 

6746 

67 

30' 

40' 

529 

772 

88 

8334 

6511 

56 

20' 

50' 

557 

805 

1.07  00 

8117 

6279 

46 

10' 

21° 

.3  584 

.3  839 

1.07  11 

2.  7904 

2.  6051 

.93  36 

69° 

cos/ 

cot/ 

CSC/ 

sec/ 

tan/ 

sin/ 

/ 

NATURAL  VALUES  OF  FUNCTIONS. 


IV 


X 

sinjr 

tanjr 

sec  A" 

cscx 

COtAf 

cosjr 

21° 

.3  584 

.3  839 

1.07  11 

2.7904 

2.  6051 

.93  36 

69° 

10' 

611 

872 

23 

7695 

5826 

25 

50' 

20' 

638 

906 

36 

7488 

5605 

15 

40' 

30' 

665 

939 

48 

7285 

5386 

04 

30' 

40' 

692 

973 

60 

7085 

5172 

.92  93 

20' 

50' 

719 

.4  006 

73 

6888 

4960 

83 

10' 

22° 

.3  746 

.4  040 

1.07  85 

2.  6695 

2.  4751 

.92  72 

68° 

10' 

773 

074 

98 

6504 

4545 

61 

50' 

20' 

800 

108 

1.08  11 

6316 

4342 

50 

40' 

30' 

827 

142 

24 

6131 

4142 

39 

30' 

40' 

854 

176 

37 

5949 

3945 

28 

20' 

50' 

881 

210 

50 

5770 

3750 

16 

10' 

23° 

.3  907 

.4  245 

1.08  64 

2.  5593 

2.  3559 

.92  05 

67° 

10' 

934 

279 

77 

5419 

3369 

.9194 

50' 

20' 

961 

314 

91 

5247 

3183 

82 

40' 

30' 

987 

348 

1.09  04 

5078 

2998 

71 

30' 

40' 

.4  014 

383 

18 

4912 

2817 

59 

20' 

50' 

041 

417 

32 

4748 

2637 

47 

10' 

24° 

.4  067 

.4  452 

1.09  46 

2.  4586 

2.  2460 

.9135 

66° 

10' 

094 

487 

61 

4426 

2286 

24 

50' 

20' 

120 

522 

75 

4269 

2113 

12 

40' 

30' 

147 

557 

89 

4114 

1943 

00 

30' 

40' 

173 

592 

1.10  04 

3961 

1775 

.90  88 

20' 

50' 

200 

628 

19 

3811 

1609 

75 

10' 

25° 

.4  226 

.4  663 

1.10  34 

2.  3662 

2.  1445 

.90  63 

65° 

10' 

253 

699 

49 

3515 

1283 

51 

50' 

20' 

279 

734 

64 

3371 

1123 

38 

40' 

30' 

305 

770 

79 

3228 

0965 

26 

30' 

40' 

331 

806 

95 

3087 

0809 

13 

20' 

50' 

358 

841 

1.11  10 

2949 

0655 

01 

10' 

26° 

.4  384 

.4  877 

1.11  26 

2.  2812 

2.  0503 

.89  88 

64° 

10' 

410 

913 

42 

2677 

0353 

75 

50' 

20' 

436 

950 

58 

2543 

0204 

62 

40' 

30' 

462 

986 

74 

2412 

0057 

49 

30' 

40' 

488 

.5  022 

90 

2282 

1.  9912 

36 

20' 

50' 

514 

059 

1.12  07 

2153 

9768 

23 

10' 

27° 

.4  540 

.5  095 

1.12  23 

2.  2027 

1.  9626 

.89  10 

63° 

10' 

566 

132 

40 

1902 

9486 

.88  97 

50' 

20' 

592 

169 

57 

1779 

9347 

84 

40' 

30' 

617 

206 

74 

1657 

9210 

70 

30' 

40' 

643 

243 

91 

1537 

9074 

57 

20' 

50' 

669 

280 

1.13  08 

1418 

8940 

43 

10' 

28° 

.4  695 

.5  317 

1.13  26 

2.  1301 

1.  8807 

.88  29 

62° 

cos/ 

cot/ 

CSC/ 

sec/ 

tan/ 

sin/ 

/ 

NATURAL  VALUES  OF  FUNCTIONS. 


X 

sinjif 

tanjf 

secx 

cscjr 

cotx 

cosx 

28° 

.4  695 

.5  317 

1.1  326 

2.1  301 

1.8  807 

.88  29 

62° 

10' 

720 

354 

343 

185 

676 

16 

50' 

20' 

746 

392 

361 

070 

546 

02 

40' 

30' 

772 

430 

379 

2.0  957 

418 

.87  88 

30' 

+0' 

797 

467 

397 

846 

291 

74 

20' 

50' 

823 

505 

415 

736 

165 

60 

10' 

29° 

.4  848 

.5  543 

1.1  434 

2.0  627 

1.8  040 

.87  46 

61° 

10' 

874 

581 

452 

519 

1.7  917 

32 

50' 

20' 

899 

619 

471 

413 

796 

18 

40' 

30' 

924 

658 

490 

308 

675 

04 

30' 

40' 

950 

696 

509 

204 

556 

.86  89 

20' 

50' 

975 

735 

528 

101 

437 

75 

10' 

30° 

.5  000 

.5  774 

1.1  547 

2.0  000 

1.7  321 

.86  60 

60° 

10' 

025 

812 

566 

1.9  900 

205 

46 

50' 

20' 

050 

851 

586 

801 

090 

31 

40' 

30' 

075 

890 

606 

703 

1.6  977 

16 

30' 

40' 

100 

930 

626 

606 

864 

01 

20' 

50' 

125 

969 

646 

511 

753 

.85  87 

10' 

31° 

.5  150 

.6  009 

1.1  666 

1.9  416 

1.6  643 

.85  72 

59° 

10' 

175 

048 

687 

323 

534 

57 

50' 

20' 

200 

088 

707 

230 

426 

42 

40' 

30' 

225 

128 

728 

139 

319 

26 

30' 

40' 

250 

168 

749 

048 

212 

U 

20' 

50' 

275 

208 

770 

1.8  959 

107 

.84  96 

10' 

32° 

.5  299 

.6  249 

1.1  792 

1.8  871 

1.6  003 

.84  80 

58° 

10' 

324 

289 

813 

783 

1.5  900 

65 

50' 

20' 

348 

330 

835 

697 

798 

50 

40' 

30' 

373 

371 

857 

612 

697 

34 

30' 

40' 

398 

412 

879 

527 

597 

18 

20' 

50' 

422 

453 

901 

443 

497 

03 

10' 

33° 

.5  446 

.6  494 

1.1  924 

1.8  361 

1.5  399 

.83  87 

57° 

10' 

471 

536 

946 

279 

301 

71 

50' 

20' 

495 

577 

969 

198 

204 

55 

40' 

30' 

519 

619 

992 

118 

108 

39 

30' 

40' 

544 

661 

1.2  015 

039 

013 

23 

20' 

50' 

568 

703 

039 

1.7  960 

1.4  919 

07 

10' 

34° 

.5  592 

.6  745 

1.2  062 

1.7  883 

1.4  826 

.82  90 

56° 

10' 

616 

787 

086 

806 

733 

74 

50' 

20' 

640 

830 

110 

730 

641 

58 

40' 

30' 

664 

873 

134 

655 

550 

41 

30' 

40' 

688 

916 

158 

581 

460 

25 

20' 

50' 

712 

959 

183 

507 

370 

08 

10' 

35° 

.5  736 

.7  002 

1.2  208 

1.7  434 

1.4  281 

.8192 

55° 

cos/ 

cot/ 

CSC/ 

sec/ 

tan/ 

sin/ 

/ 

NATURAL  VALUES  OF  FUNCTIONS. 


VI 


X 

sin;r 

tanx 

secAf 

CSCA" 

cotjr 

COSJf 

"~~ 

35° 

.5  736 

.7  002 

1.2  208 

1.7  434 

1.4  281 

.8192 

55° 

10' 

760 

046 

233 

362 

193 

75 

50' 

20' 

783 

089 

258 

291 

106 

58 

40' 

30' 

807 

133 

283 

221 

019 

41 

30' 

40' 

831 

177 

309 

151 

1.3  934 

24 

20' 

50' 

854 

221 

335 

081 

848 

07 

10' 

36° 

.5  878 

.7  265 

1.2  361 

1.7  013 

1.3  764 

.80  90 

54° 

10' 

901 

310 

387 

1.6  945 

680 

73 

50' 

20' 

925 

355 

413 

878 

597 

56 

40' 

30' 

948 

400 

440 

812 

514 

39 

30' 

40' 

972 

445 

467 

746 

432 

21 

20' 

50' 

995 

490 

494 

681 

351 

04 

10' 

37° 

.6  018 

.7  536 

1.2  521 

1.6  616 

1.3  270 

.79  86 

53° 

10' 

041 

581 

549 

553 

190 

69 

50' 

20' 

065 

627 

577 

489 

111 

51 

40' 

30' 

088 

673 

605 

427 

032 

34 

30' 

40' 

111 

720 

633 

365 

1.2  954 

16 

20' 

50' 

134 

766 

661 

303 

876 

.78  98 

10' 

38° 

.6  157 

.7  813 

1.2  690 

1.6  243 

1.2  799 

.78  80 

52° 

10' 

ISO 

860 

719 

183 

723 

62 

50' 

20' 

202 

907 

748 

123 

647 

44 

40' 

30' 

225 

954 

778 

064 

572 

26 

30' 

40' 

248 

.8  002 

807 

005 

497 

08 

20' 

50' 

271 

050 

837 

1.5  948 

423 

.77  90 

10' 

39° 

.6  293 

.8  098 

1.2  868 

1.5  890 

1.2  349 

.77  71 

51° 

10' 

316 

146 

898 

833 

276 

53 

50' 

20' 

338 

195 

929 

777 

203 

35 

40' 

30' 

361 

243 

960 

721 

131 

16 

30' 

40' 

383 

292 

991 

666 

059 

.76  98 

20' 

50' 

406 

342 

1.3  022 

611 

1.1  988 

79 

10' 

40° 

.6  428 

.8  391 

1.3  054 

1.5  557 

1.1  918 

.76  60 

50° 

10' 

450 

441 

086 

504 

847 

42 

50' 

20' 

472 

491 

118 

450 

778 

23 

40' 

30' 

494 

541 

151 

398 

708 

04 

30' 

40' 

517 

591 

184 

345 

640 

.75  85 

20' 

50' 

539 

642 

217 

294 

571 

66 

10' 

41° 

.6  561 

.8  693 

1.3  250 

1.5  243 

1.1  504 

.75  47 

49° 

10' 

583 

744 

284 

192 

.  436 

28 

50' 

20' 

604 

796 

318 

141 

369 

09 

40' 

30' 

626 

847 

352 

092 

303 

.74  90 

30^ 

40' 

648 

899 

386 

042 

237 

70 

20' 

50' 

670 

952 

421 

1.4  993 

171 

51 

10' 

42° 

.6  691 

.9  004 

1.3  456 

1.4  945 

1.1  106 

.74  31 

48° 

cos/ 

cot/ 

CSC/ 

sec/ 

tan/ 

sin/ 

/ 

Vll 


NATURAL    VALUES   OF   FUNCTIONS. 


X 

sin  X 

tan  jr 

sec  x 

CSC  X 

cot  Jf 

cos  jr 

42° 

.6  691 

.9  004 

1.3  456 

1.4  945 

1.1  106 

.74  31 

48° 

10' 

713 

057 

492 

897 

041 

12 

50' 

20' 

734 

110 

527 

849 

1.0  977 

.73  92 

40' 

30' 

756 

163 

563 

802 

913 

73 

30' 

40' 

777 

217 

600 

755 

850 

53 

20' 

50' 

799 

271 

636 

709 

786 

ZZ 

10' 

43° 

.6  820 

.9  325 

1.3  673 

1.4  663 

1.0  724 

.73  14 

47° 

10' 

841 

380 

711 

617 

661 

.72  94 

50' 

20' 

862 

435 

748 

572 

599 

74 

40' 

30' 

884 

490 

786 

527 

538 

54 

30' 

40' 

905 

545 

824 

483 

477 

34 

20' 

50' 

926 

601 

863 

439 

416 

14 

10' 

44° 

.6  947 

.9  657 

1.3  902 

1.4  396 

1.0  355 

.7193 

46° 

10' 

967 

713 

941 

352 

295 

73 

50' 

20' 

988 

770 

980 

310 

235 

53 

40' 

30' 

.7  009 

827 

1.4  020 

267 

176 

33 

30' 

40' 

030 

884 

061 

225 

117 

12 

20' 

50' 

050 

942 

101 

183 

058 

.70  92 

10' 

45° 

.7  071 

1.0  000 

1.4  142 

1.4  142 

1.0  000 

.70  71 

45° 

cos/ 

cot/ 

CSC/ 

sec/ 

tan/ 

sin/ 

/ 

Circular 
Measure. 


{»= 


X 

180° 


Hyperbolic/  sinh  u 
Functions.    I  cosh  u 


tanx. 
sec  jr. 


X 

e 

X 

e 

X 

u 

X 

u 

1° 

.01745 

V 

.00029 

2° 

.0349 

40° 

.7629 

2° 

.03491 

2' 

.00058 

4° 

.0699 

42° 

.8092 

3° 

.05236 

3' 

.00087 

6° 

.1049 

44° 

.8569 

4° 

.06981 

4' 

.00116 

8° 

.1401 

46° 

.9063 

5° 

.08727 

5' 

.00145 

10° 

.1754 

48° 

.9575 

6° 

.10472 

6' 

.00175 

12° 

.2110 

50° 

1.0107 

7° 

.12217 

7' 

.00204 

14° 

.2468 

52° 

1.0662 

8° 

.13963 

8' 

.00233 

16° 

.2830 

54° 

1.1242 

9° 

.15708 

9' 

.00262 

18° 

.3195 

56° 

1.1851 

10° 

.17453 

10' 

.00291 

20° 

.3564 

58° 

1.2492 

20° 

.34907 

20' 

.00582 

22° 

.3938 

60° 

1.3170 

30° 

.52360 

30' 

.00873 

24° 

.4317 

62° 

1.3890 

40° 

.69813 

40' 

.01164 

26° 

.4702 

64° 

1.4659 

50° 

.87266 

50' 

.01454 

28° 

.5094 

66° 

1.5485 

60° 

1.04720 

60' 

.01745 

30° 

.5493 

68° 

1.6379 

70° 

1.22173 

32° 

.5900 

70° 

1.7354 

80° 

1.39626 

10" 

.00005 

34° 

.6317 

72° 

1.8427 

90° 

1.57080 

20" 

.00010 

36° 

.6743 

74° 

1.9623 

100° 

1.74533 

40" 

.00019 

38° 

.7180 

76° 

2.0973 

ELEMENTS    OF    TRIGONOMETRY. 


j;04c 


INTRODUCTION. 

Trigonometry  investigates  the  numerical  relations  be- 
tween segments,*  angles,  and  areas  of  figures. 

1.  The  meaning  of  geometric  operations  will  be  retained  in 
trigonometry.  Thus,  Fig.  1,  if  the  segment  CD,  of  length 
a,  is  added  to  the  segment  DE, 
of  length  h,  the  sum  is  the  seg- 
ment CE,  of  length  a  -f  6. 

In  Fig.  2,  if  DE,  of  length  h, 
is  subtracted  from  CD,  of  length 
a,  the  remainder  CE  is  of  length 
a  —  b. 

In  Fig.  3,  if  CD  is  of  length  a,      ^        ^    yig.  3. 
CE  =  a-}-a-fa-|-a  =  4a. 

In  Fig.  4,  if  CD  is  of  length  a,  CE  (two  of  the  ten  equal 
parts  of  CD)  is  of  length  .2  a. 

Hence  if  a  segment  .  g 

of  length  a  is  mul-  C  E  D 

tiplied    by   2.4,    the  Fig.  4. 

product  will  be  considered  either  a  divided  by  10  and  the 

*  In  this  work  a  straight  line,  or  simply  a  line,  is  conceived  to  be 
infinitely  long  in  two  opposite  directions  from  a  point ;  hence  a  half- 
line  is  conceived  to  be  infinitely  long  in  one  direction  from  a  point ; 
and  a  segment  of  a  line,  or  simply  a  segment,  is  the  name  which  will 
be  applied  to  a  limited  portion  of  a  line. 

1 


1           fl 

b 

C                              D 
Fig.  1. 

1                         0 

E 

C              E            b 
Fig.  2. 

.     a    .    a     .     a     . 

n 

D 

C         D 

E 

2  INTRODUCTION. 

quotient  multiplied  by  24,  or  a  multiplied  by  24  and 
the  product  divided  by  10. 

In  Fig.  5,  EF,  of  length  h,  may  be  conceived  as  being 
divided  into  CD,  of  length  a.  If  CD  contains  EF  five  times, 
the  ratio   of   CD  to  £  p 

EF  is  said  to  be  5.   ' — ' 

If   it   is   desired   to   ,  _    a  _  _      , 

obtain   the   ratio   of  ^     G  D 

Fir     5 

EF  to    CD,    CD  is 

first  divided  into  10  equal  parts  and  one  of  these  parts 

CG  applied  to  EF.     If  EF  contains  CG  exactly  two  times, 

the  ratio  of  EF  to  CD  is  said  to  be  .2.     The  ratios  -  =  5 

b  ^ 

and  -  =  .2  have  a  product  of  1,  and  are  said  to  be  reciprocal 

ratios. 

In  Fig.  6,  EF  is  not  contained  exactly  in  CD,  but  is  con- 
tained twice  and  a  remainder  GD  more  than  three-tenths  of 
EF,  but  less  than  three      ,  ■■.■.....  , 

and  one-half  tenths ;  the      ^ ^ ,_,_p-^ 

ratio  of   CD  to  EF  is     C  G       D 

said  to  be  2.3+.     This  F^^-  ^■ 

value  of  the  ratio  contains  an  error  not  greater  than  .05. 
The  ratio  of  the  error  to  the  assumed  value  is  not  greater 
than  .05  in  2.3,  or  5  in  230,  or  1  in  46.  If  this  error  in  2.3, 
the  assumed  value  of  the  ratio,  is  too  great  for  use  in  a 
problem  in  which  the  ratio  enters,  EF  is  divided  into  100, 
1000,  10,000,  etc.,  equal  parts,  and  one  of  them  applied  to 
GD.  Thus,  if  EF  is  divided  into  10,000  parts  and  GD  is 
found  to  contain  one  of  these  parts  not  less  than  3470.5 
times,  but  less  than  3471  times,  the  ratio  of  CD  to  EF  is 
said  to  be  2.3471 -.  This  value  of  the  ratio  contains  an 
error  not  greater  than  .00005.  The  ratio  of  the  error  to  the 
assumed  value  is  not  greater  than  .00005  in  2.3471,  or  1  in 
46,942. 

In  the  same  way,  if  the  ratio  of  CD  to  EF  is  desired,  CD 


NEGATIVE   SEGMENTS  AND  ANGLES.  3 

is  divided  into  10,  100,  1000,  10,000,  etc.,  equal  parts,  and 
so  divided  applied  to  EF.  Suppose  CD  is  divided  into 
10,000  parts,  and  the  beginning  of  CD  is  placed  at  E. 
Then  if  the  4261st  division  of  CD  is  the  one  nearest  to  F, 
the  ratio  of  EF  to  CD  is  said  to  be  approximately  .4261 ; 
the  ratio  of  the  error  to  the  assumed  value  of  the  ratio  in 
this  case  is  not  greater  than  1  in  8522. 

Operations  upon  angles.  —  All  that  has  been  said  with 
reference  to  addition,  subtraction,  multi- 
plication, and  division  of  segments  will 
apply  to  corresponding  operations  per- 
formed upon  angles.  Thus,  Fig.  7,  the 
sum  of  angle  BAC{=a)  and  angle 
CAD  (=  6)  is  the  angle  BAD(=  a  +  b).  Fig. 

Similarly,  the  difference  of  BAC(=a) 
and  CAD(—b),  Fig.  8,  is  the  angle 
BAD(=a-b). 

In  a  similar  manner,  the  interpreta- 
tion of  the  multiplication  of  an  angle  by 
a  number,  the  division  of  an  angle  by  a 
number,  and  the  division  of  an  angle  by  ^^'  ^' 

another  angle  will  be  conceived  to  be  the  same  as  in 
geometry. 

NEGATIVE  SEGMENTS  AND  ANGLES. 

2.   Assume  the  direction  of  measurement  of  the  segment 

CD,  of  length  a,  Fig.  9,  to  be  from  ^         

left   to  right.     From  D,  the  ter-   E  C     «     5 

minal  point  of  CD,  lay  off  a  seg-  ^^^'  ^' 

ment  DE,  of  length  b,  backward  on  CD.  Then  if  b  is 
greater  than  a,  E  will  be  to  the  left  of  C,  but  a  —  b  will 
still  be  said  to  be  the  length  of  the  segment  GE,  measured 
from  C  to  E,  or  in  the  direction  opposite  to  that  assumed 
to  be  positive.     Now  the  length  of  a  in  the  discussion  is 


4  INTRODUCTION. 

arbitrary,  and  may  be  conceived  to  be  equal  to  the  length 
of  CD  in  Fig.  10,  or  zero ;  in  which  case  the  segment  CE 

(measured   from   right  to   left,   or 5 p 

opposite   the  direction  assumed  to    ^  C 

be  positive)  is  equal  to  —  h.  ^^^'  ^^' 

Hence,  if  any  segment  is  multiplied  by  —  1,  the  effect  of 
the  operation  is  to  measure  the  segment  from  the  same  point 
as  at  first  but  in  an  opposite  direction ;  thus,  —  1  x  a  =  —a, 
and  — Ix  —  a=  +  a. 

The  multiplication  of  a  segment  by  a  positive  number  will 
be  considered  to  be  repetition  of  the  segment  in  the  same 
direction  the  number  of  times  indicated  by  the  multi- 
plier (which  may  be  either  integral  or  fractional) ;  thus, 
4.3  times  —  a  =  —  (4.3  a).  The  product  of  a  segment  by  a 
negative  number  will  be  conceived  to  be :  first,  a  reversal 
of  the  direction  of  the  segment,  and  then  the  multiplication 
by  a  positive  number ;  thus, 

-  4.3  X  a  =  4.3  X  -  1  X  a  =  4.3  X  -  a  =  -(4.3  a). 
Also 

-4.3  x-a  =  4.3  x-1  x-a  =  4.3  xa  =  +  (4.3a). 

If  division  is  defined  as  being  the  process  of  obtaining  a 
quantity,  which,  multiplied  with  the  divisor,  will  produce 
the  dividend,  each  process  of  multiplication  gives 

-(4-3 «)   ■      4.3.  3(i3a)  =  _„. 

—  a  4.3 

-(4-3  g)^      ^3  -(4.3  g)^     ^ 

a  '  '  -4.3 

+  (4.3  g)^      ^.3.  ±M^=:-a. 

—  a  —  4.3 

The  same  methods  of  reasoning  will  give  the  conception 
of  negative  angles.  Thus,  if  turning  a  screw  in  one  direction 
is  positive  turning,  then  turning  the  screw  in  the  opposite 
direction  is  negative  turning,  because  it  cancels  or  undoes 


PRODUCTS  AND   QUOTIENTS   OF   SEGMENTS.  5 

an  equal  amount  of  positive  turning ;  and  negative  angles  so 
conceived  may  be  multiplied  by  positive  or  negative  num- 
bers, or  divided  by  positive  or  negative  numbers  or  angles. 


PRODUCTS  AND  QUOTIENTS  OF  SEGMENTS. 

3.  In  geometry,  the  product  of  two  segments  is  explained 
to  be  a  term  used  to  aid  us  in  counting  the  number  of  units 
of  area  in  a  surface ;  thus  if  a  rectangle  has  sides  of  length 
3  inches  and  2  inches,  respectively,  the  area  is  said  to  be 
3  inches  x  2  inches  =  6  square  inches.  This  statement  is 
a  short  way  of  saying  that  in  the  rectangle  there  can  be 
arranged  2  rows  of  inch-squares,  3  in  a  row,  making  a 
total  of  6  inch-squares  or  6  square  inches.  In  the  same 
way  the  operation  4  inches  x  3  inches  x  2  inches  =  24  cubic 
inches,  is  a  method  of  counting  cubes  in  a  rectangular  par- 
allelopiped.  In  trigonometry,  the  same  meaning  is  attached 
to  these  expressions  as  in  geometry.  Thus,  the  product  of 
a  segment  and  an  abstract  number,  say  4.3  a,  is  said  to  be  a 
length;  the  product  (so  called)  of  two  segments  and  an 
abstract  number,  say  4.3  a^  or  4.3  ah,  is  said  to  be  an  area ; 
the  product  of  three  segments  and  an  abstract  number,  say 
4.3  ahc  or  4.3  a^h,  is  said  to  be  a  volume. 

Since  division  is  the  inverse  of  multiplication,  an  area  is 

indicated  by  the  quotient  of  a  volume  by  a  segment,  such 

4.3 «6c       C5-     -1    1       4.3  a5c    4.3 a&       /a  o    k      ^/TWTTT^ 

represent  segments. 

Expressions  having  no  geometric  meaning.  —  In  geometry, 
expressions  for  lengths,  areas,  and  volumes  have  meaning ; 
but  many  forms  which  occur  in  trigonometric  reductions 

such  as  Va,  — ,  -^abc,  4.3  abed,  are  said  to  have  no  geo- 
a 

metric  meaning. 


6  INTRODUCTION. 

Dimensions.  —  Any  expression  representing  a  length  is 
said  to  be  of  one  dimension ;  an  expression  representing  an 
area  is  said  to  be  of  two  dimensions ;  and  any  expression 
is  of  h  dimensions  if  the  algebraic  sum  of  the  exponents  of 
the  segments  in  each  term  is  k.     Thus  a  ratio  has  zero 

a  2 

dimensions,  since  it  is  of  the  form   -  =  a}h~'^ ;    -  has  —  1 

h  a 


dimensions ;   -yob  =  a^b^  is  of  |-  dimensions ;  abed  is  of  4 
dimensions ;  and  so  on. 

4.  Homogeneity  of  trigonometric  equations.  —  No  geometric 
meaning  can  be  attached  to  the  result  obtained  by  adding 
an  area  and  a  length ;  or  by  adding  a  volume  and  a  length ; 
or  by  adding  a  volume  and  an  area.  And  any  equation 
involving  terms  of  different  dimensions  is  also  without 
meaning.  Hence  in  every  trigonometric  equation,  each 
term  must  have  the  same  dimensions.  Equations  in  which 
this  condition  is  fulfilled  are  called  homogeneous  equations. 
Therefore  every  trigonometric  equation  must  be  a  homo- 
geneous equation. 

This  property  of  trigonometric  equations  is  frequently 
of  value  as  a  check  on  the  correctness  of  reduction  pro- 
cesses. For  example,  suppose  an  investigation  of  the  rela- 
tions of  the  parts  of  any  figure  is  undertaken.  The  first 
equation  used  must  be  homogeneous ;  any  other  equation 
derived  from  it  should  also  be  homogeneous ;  if  it  fails  in 
homogeneity  a  mistake  has  been  made  in  the  process  of 
reduction. 

Exercises.  —  1.  What  are  the  dimensions  of  each  of  the 
following  expressions : 


3  7  a'bc         3  ab&     a^U       1  l 

2.  Write  an  equation  containing  six  terms  of  different 
form,  each  term  representing  a  segment ;  also  an  equation 
containing  six  terms,  each  representing  an  area. 


APPROXIMATE   VALUES  OF  NUMBERS.  7 

5.  Significant  figures.  —  Zeros  are  not  significant  figures, 
if  they  are  supplied  to  a  number  merely  for  the  purpose  of 
putting  the  decimal  point  in  the  right  position.  Thus  in  the 
number  .00908,  the  two  zeros  immediately  following  the 
decimal  point  are  not  significant  figures,  but  the  remaining 
figures,  9,  0,  and  8,  are  significant  figures. 

Also,  in  writing  approximate  values  of  large  numbers, 
zeros  are  frequently  supplied  when  necessary  to  get  the 
first  significant  figure  of  the  number  in  the  proper  position 
with  reference  to  the  unit's  place.  Thus  in  the  number 
92,500,000  miles,  the  5  is  understood  to  be  the  nearest  fig- 
ure in  the  hundred-thousands  place;  the  zeros  following 
the  5  indicate  a  lack  of  knowledge  as  to  the  value  of  the 
figures  in  the  ten-thousands  place  and  in  lower  orders ;  the 
9,  2,  and  5  are  the  only  significant  figures  in  this  number. 

6.  Errors  in  approximate  values  of  numbers.  —  Suppose  an 
assumed  value  of  a  number,  expressed  with  four  significant 
figures,  the  fourth  figure  being  the  nearest  figure  in  that 
place.  The  ratio  of  the  error  to  the  assumed  value  of  the 
number  is  not  greater  than  1  to  2000.  For  the  most  un- 
favorable case  is  when  1000  is  assumed  as  the  value  of 
1000.5,  in  which  case  the  ratio  is  1  to  2000. 

In  a  similar  manner  the  ratio  of  the  error  of  an  approxi- 
mate value  of  a  number  of  five  significant  figures  (the  fifth 
figure  being  the  nearest  figure)  cannot  be  greater  than  1  to 
20,000;  and  soon. 

7.  Errors  arising  in  the  processes  of  calculation.  —  Suppose 
that  assumed  values  of  numbers  are  introduced  into  a 
numerical  calculation.  From  the  principles  of  arithmetic 
all  numbers  arising  subsequently  in  the  process  of  calcula- 
tion will  in  general  have  about  the  same  degree  of  precision 
as  was  contained  in  the  assumed  value  of  the  least  degree  of 
precision  which  entered  into  the  calculation.  Thus,  con- 
sider the  following  examples : 


8  INTRODUCTION. 

Multiplication.  —  Suppose  it  is  desired  to  multiply  the 
approximate  value  .7862  by  the  number  43.59.  The  num- 
ber of  which  .7862  is  the  approximate  value  is  between 
.78615  and  .78625.  Multiply  each  separately  by  43.59. 
The  products  are  34.2682785  and  34.2726375.  It  will  be 
seen  that  34.27  will  represent  either  product  to  four  signifi- 
cant figures.  Now  multiply  .7862  by  43.59  by 
.7862  first  multiplying  by  40,  then  by  3,  then  by  .5, 

43.59  then  by  .09. 

To  reject  all  figures  to  the  right  of  the 
vertical  line  drawn  through  the  products,  use 
all  the  figures  of  the  multiplicand  when  multi- 
plying by  40,  and  put  in  the  decimal  point ;  in 
multiplying  by  3,  strike  out  the  2  of  the  multi- 
plicand, but  use  it  to  obtain  figures  which  may 
affect  the  column  to  the  left  of  the  vertical 
line ;  thus,  3  times  2  are  6,  and  6  being  more 
than  5,  1  is  carried ;  3  times  6  are  18,  and  1  to 
carry  is  19,  and  so  on.  In  multiplying  by  5, 
the  6  of  the  multiplicand  is  struck  out,  but  is 
used  to  obtain  3  to  carry ;  thus,  5  times  6  are 
30,  5  times  8  are  40,  and  3  to  carry  are  43,  and 
34.27  so  on.     In  multiplying  by  9,  the  8  is  struck 

out,  but  the  86  now  struck  out  being  nearly  90, 
we  proceed :  9  times  6  are  54,  9  times  8  are  72,  and  5  to 
carry  are  77,  9  times  7  are  63,  and  8  to  carry  are  71 ;  or 
simply  considering  the  86  thrown  away  to  be  90,  we  say  9 
times  9  are  81,  9  times  7  are  63,  and  8  to  carry  are  71.  In 
adding  the  partial  products,  the  right-hand  column  adds  to 
21 ;  reject  the  1  and  carry  the  2  to  the  next  column,  as  only 
four  significant  figures  of  the  product  can  be  correctly 
obtained.     The  method  is  expressed  in  the  following 

Rule.  —  Use  the  figures  of  the  multiplier  in  reverse  order. 
Supply  the  decimal  point  in  the  first  partial  product.  Eeject 
all  unnecessary  figures. 


31.448 

0 

2.358 

6 

.393 

10 

.070 

758 

34.270 

458 

.7862 

43.59 

31.448 

0 

2  359 

393 

71 

CONTKACTIONS  IN  COMPUTATIONS. 


9 


Division.  —  To   reverse   the   process,  divide  the   product 
34.27  by  78.62. 


COMMON   METHOD. 


CONTRACTED    METHOD. 


7862)34.27 
3144 

00000(43.589 
8 

.7862)34.27 
3145 

2  82 
2  36 

46 
39 

"7 

2  82 
2  35 

46 
39 

20 
S6 

340 
310 

7 
6 

0300 
2896 

74040 
70758 

(43.59 


Crosses  are  added  to  make  the  number  of  decimal  places 
in  the  dividend  equal  to  the  number  in  the  divisor.  After 
the  multiplication  by  the  quotient  figure  4,  put  as  many 
crosses  after  the  4  as  there  are  unused  crosses  in  the  divi- 
dend. Then  supply  the  decimal  point.  The  rest  of  the 
process  is  similar  to  the  process  of  multiplication. 

The  square  root  of  .7862. 


COMMON    METHOD. 

.78620000(.88667 
64 


CONTRACTED    METHOD. 


168 

1462 
1344 

1766 

118 
105 

00 
96 

17726 

12 
10 

0400 
6356 

17732 

1 

4044 

.7862(.8867 
64 


168 


176 


1462 
1344 


118 
106 

12 


10  INTRODUCTION. 

A  consideration  of  these  examples  will  show  that  if  data 
are  correct  to  only  three  significant  figures  (say  an  error  of 
1  in  1000),  the  calculation  may  be  kept  in  four  significant 
figures ;  for  the  result  will  be  as  precise  as  the  data,  and 
further  figures  are  unnecessary  and  unreliable. 

Similarly,  if  the  data  are  correct  to  four  significant  figures 
(say  an  error  of  1  in  10,000),  the  calculation  may  be  kept  in 
five  significant  figures.     And  so  on. 

Applications.  —  The  summation 
of  series  is  one  of  the  many  prac- 
tical applications  of  approximate 
values  of  numbers.  Thus,  sup- 
pose it  is  desired  to  compute  to 
ten  decimal  places  the  value  of 
the  series : 


1) 

1. 

2) 

1. 

3) 

.5 

4) 

.166,666,666,67 

5) 

41,666,666,67 

6) 

8,333,333,33 

7) 

1,388,888,89 

8) 

198,412,70 

9) 

24,801,59 

10) 

2,755,73 

11) 

275,57 

12) 

25,05 

13) 

2,09 

14) 

16 

1 

2.718,281,828,46 

e  =  1  -h  -  +  — —  H = h  etc., 

1      1.2     1.2-3  ' 

in  which  each  term  is  obtained 
from  the  one  preceding  by  divi- 
sion, the  divisors  increasing  by  1 
each  time.  The  work  proceeds  as 
in  the  margin.  It  will  be  seen 
that  the  nearest  tenth  decimal 
figure  is  5. 

Problems. — Perform  the  following  multiplications,  keep- 
ing only  the  figures  which  affect  the  fourth  significant  figure 
of  the  product.  After  each  multiplication,  perform  the  re- 
verse operation  of  division  or  square  root : 

234.1  X  4.444  =  1040  1.414  x  1.414  =  2.000 

.4567  X  9.009  =  4.114  .7454  x  .7454  =  .5000 

7060  X  .0302  =  213.2  57.33  x  57.33  =  3287 

.7854  X  34.25  =  26.90  1.012  x  1.012  =  1.024 


CHAPTER  I. 


FUNCTIONS  OF  ANGLES. 

8.  Generation  of  an  angle.  —  The  angle  BAC  may  be  con- 
ceived to  be  generated  by  a  turning  of  a  half-line  from 
either   side  of  the   angle  q 

to  the  other  side.  Thus, 
the  position  AC  may  be 
reached  by  a  turning  of  a 
half-line  from  AB  through 
an  angle  x  +  any  num- 
ber of  revolutions;  or 
the  same  position  may 
be  reached  by  a  turning 

from  AB  in  an  opposite  direction  through  an  angle 
360°  —  a;  -f  any  number  of  revolutions.  Also,  if  b,  the 
length  of  AC,  is  regarded  as  negative,  the  turning  from 
AB  is  either  180° -|- a;  or  a  downward  turning  of  180°— .t. 
For  example,  if,  in  Fig.  11,  x  is  30°  and  b  is  10  inches,  the 
point  C  is  determined  by  any  of  the  following  values  of  b 
and  X : 

6  =  -1-10;  a;  =  30°,     390°,  750°,  1110°,  etc. 

b  =  -10;  x  =  210°,  570°,  930°,  1290°,  etc. 

6  =  +  10;  a;  =-330°,  -690°,  -1050°,  etc. 

&  =  -10;  a;  =  -150°,  -510°,  -870°,  etc. 

The  half-line  AB  from  which  angles  will  be  conceived  to 
be  measured  will  be  called  the  initial  line ;  the  final  position 
AC  of  the  revolving  half-line  will  be  called  the  terminal 

11 


12 


FUNCTIONS  OF  ANGLES. 


line  of  the  angle ;  the  length,  b,  from  the  vertex  of  the  angle 
to  any  point  C  on  the  terminal  line  will  be  called  the 
distance  of  C. 

The  direction  of  the  initial  line  is  entirely  arbitrary.  It 
will  be  drawn  to  the  right  in  the  discussions  in  this  chapter. 
The  angle  will  be  regarded  as  positive  when  generated  by  a 
half-line  turning  from  the  initial  line  upward  to  the  left, 
as  indicated  by  the  arrow,  Fig.  11.  It  will  be  regarded  as 
negative  when  generated  by  a  half-line  turning  in  the  oppo- 
site direction.  The  distance  will  be  regarded  as  positive 
when  laid  off  from  the  vertex  of  the  angle  along  the 
terminal  line. 


9.  The  four  quadrants.  —  When  a  positive  angle  is  under 
consideration,  it  will  be  conceived  to  be  moved  so  that 
its  vertex  is  brought  at  A, 
one  side  coincides  with  AB, 
and  the  angle  extends  in  a 
positive  direction.  Then  if 
the  teitoiinal  line  falls  in 
the  angle  BAD  (where  DDi 
is  perpendicular  to  BABi), 
the  angle  is  said  to  be  of 
the  first  quadrant;  if  the 
terminal  line  falls  in  DAB^, 
the  angle  is  of  the  second 
quadrant;  if  the  terminal 
line  is  in  BiAD^,  the  angle 
is  of  the  third  quadrant ;  if  the  terminal  line  is  in  DiAB, 
the  angle  is  of  the  fourth  quadrant.     Thus, 

30°,  390°,  -330°,  -690°  are  angles  of  the  first  quadrant. 

150°,  510°,  -210°,  -570°  are  angles  of  the  second  quadrant. 

210°,  570°,  -150°,  -510°  are  angles  of  the  third  quadrant. 

330°,  690°,  -  30°,  -390°  are  angles  of  the  fourth  quadrant. 


DEFINITION  OF  FUNCTION. 


13 


c 

Fig.  13. 


Abscissa  and  ordinate.  — From  any  point  C  drop  a  perpen- 
dicular upon  tlie  initial 
line  AB.  The  distance 
from  A  to  the  foot  of 
this  perpendicular  is 
called  the  abscissa  of 
C.  The  length  of  the 
perpendicular  is  called 
the  ordinate  of  C. 

Abscissae  measured  to 
the  right  of  A  will  be 
considered  positive,  and 
measured  to  the  left, 
negative.  Ordinates 

measured  upward  from 
the  initial  line  (or  ini- 
tial line  drawn  back- 
ward) will  be  considered 
positive,  and  measured 
downward,        negative.  ^ 

Thus,  Fig.  14. 


C, 


In  Quadrant     I,  a  is  +  and  c  is  -h 
In  Quadrant    II,  a  is  +  and  c  is  — . 
In  Quadrant  III,  a  is  —  and  c  is  — . 
In  Quadrant  IV,  a  is  —  and  c  is  +. 

On  the  half-line  AB,  a  is  zero  and  c  is  -h. 
On  the  half-line  AD,  a  is    +    and  c  is  zero. 
On  the  half-line  AB^^,  a  is  zero  and  c  is  — . 
On  the  half-line  AD^,  a  is    —    and  c  is  zero. 

Definition  of  function.  —  Suppose  a  magnitude  m,  and  a 
number  of  other  magnitudes,  p,  q,  r,  so  related  to  m,  that 
when  m  changes  in  value,  all  of  the  others  change  in  value ; 


14 


FUNCTIONS   OF  ANGLES. 


and  when  m  remains  fixed  in  value,  all  of  the  others  remain 
fixed.     In  such  a  case,  p,  q,  r  are  said  to  be  functions  of  m. 

10.   The  six  ratios.  —  If  we  take  in  pairs  the  ordinate, 
distance,  and  abscissa  of  a  point  on  the  terminal  line  of  an 
angle  A,  six  different  ratios  can  be  obtained : 
a     a     h     h    c     c 


It  will  now  be  shown  that  these  six  ratios  are  functions  of 
the  angle  A. 

For  suppose,  first,  that  h  remains  the  same  length  while 
the  angle  A  changes  in 
magnitude ;  say  A  in- 
creases. Then  if  A  is  an 
angle  of  the  first  quadrant, 
the  point  C  will  move 
upward  to  the  left,  its  or- 
dinate will  increase,  and 
its  abscissa  will  decrease. 

Hence  -,  -,  and  -  will  in- 
h  c  c 

crease,  and   their  reciprocal  ratios  will  decrease. 

In  a  similar  manner  by  assuming  a  or  c  to  remain  the 
same  length,  or  by  assuming  the  moving  point  C  to  be  in 
any  other  quadrant,  the  same  thing  may  be  shown. 

Assume  now  that  C  moves  so  that  the  angle  A  is  un- 
changed. If  C  moves  either  toward  A  or  away  from  it  to 
a  point  C,  the  similar  tri-  . 


angles   ABC    and   AB'C 

,.r 

give: 

>-'"■'    ■■ 

BC:AC  =  B'C':AC' 

A^ 

y^ 

BC:AB  =  B'C':AB' 

^ 

a 

AC:AB=  AC  :  AB'           ^ 

c 

the  reciprocal  ratios  also   A 

I 

J' 

B" B 

being  equal  in  pairs. 

Fig. 

16. 

NAMES  OF  THE   RATIOS. 


15 


Names  of  the  ratios.  — For  any  point  on  the  terminal  line 
of  an  angle, 

The  ratio  of  the  ordinate  to  the  distance  is  called  the 
sine  of  the  angle.     The  sine  of  x  is  written  sin  a;. 

The  ratio  of  the  ordinate  to  the  abscissa  is  called  the 
tangent  of  the  angle.     The  tangent  of  x  is  written  tan  x. 

The  ratio  of  the  distance  to  the  abscissa  is  called  the 
secant  of  the  angle.     The  secant  of  x  is  written  sec  x. 

These  three  ratios  connect  the  ordinate,  distance,  and 
abscissa  of  a  point ;  but  it  is  often  convenient  to  use  the 
reciprocal  ratios. 

Imagine  the  triangle  ABCi  to  be  taken  up  from  the  plane 
and  placed  so  that  the  vertex  C'l  falls  at  A,  and  the  side 
CiB  coincides  with  the  initial  line.     B  takes  the  position 


B\  and  A  the  position  A'.  Then  the  ordinate  of  A'  will 
equal  the  abscissa  of  (7i,  and  the  abscissa  of  A'  will  equal 
the  ordinate  of  C^.     The  three  reciprocal  ratios  are : 


16  FUNCTIONS  OF  ANGLES. 

First.  The  ratio  of  the  distance  of  Ci  (distance  of  A)  to 
the  ordinate  of  Oi  (abscissa  of  J.')  is  the  secant  of  B'AA' 
(=C,  =  y). 

Second.  The  ratio  of  the  abscissa  of  Cy  (ordinate  of  A') 
to  the  ordinate  of  (7i  (abscissa  of  A')  is  the  tangent  of  y. 

Tliird.  The  ratio  of  the  abscissa  of  C]  (ordinate  of  A^)  to 
the  distance  of  (7i  (distance  of  A')  is  the  sine  of  y. 

Now  the  angle  Cj  (=BAA'  =  y)  is  the  complement  of 
the  angle  BACi  (=x);  that  is,  y  =  90°—  x;  on  this  account 
any  function  of  y  is  called  the  function  of  the  complement 
of  X  or  the  complement's  function,  or  simply  cof unction  of  x.* 

Although  the  prefix  co  was  obtained  by  thinking  of  y, 
and  therefore  x  as  being  acute,  the  functions  of  y  or  the 
cof  unctions  of  x  will  be  defined  as  applying  to  all  angles. 

The  co-named  functions.  —  For  any  point  on  the  terminal 
line  of  an  angle : 

The  ratio  of  the  distance  to  the  ordinate  is  called  the 
cosecant  of  the  angle.     The  cosecant  of  x  is  written  esc  x. 

The  ratio  of  the  abscissa  to  the  ordinate  is  called  the 
cotangent  of  the  angle.     The  cotangent  of  x  is  written  cot  x. 

The  ratio  of  the  abscissa  to  the  distance  is  called  the 
cosine  of  the  angle.     The  cosine  of  x  is  written  cosic. 

When  the  group  of  functions,  esc  x,  cot  x,  cos  x,  are 
referred  to,  they  will  be  spoken  of  as  the  co-named  func- 
tions of  X. 

But  it  must  be  noticed  that  sin  a;,  tancc,  secx  are  cof  unc- 
tions ;  that  is,  sin  x  is  the  cofunction  of  cos  x,  just  as  cos  x 
is  the  cofunction  of  since.  In  the  same  way  tanic  and 
cotic  are  cofunctions  of  each  other;  secx  and  cscic  are 
also  cofunctions  of  each  other. 

*  Note  that  the  complement  of  the  complement  of  an  angle  is  the 
angle  itself  [90°  —  (90°  —  x)  =  x] ,  and  that  any  co-cof unction  of  the 
angle  would  be  the  function  itself. 


RELATIONS  BETWEEN  FUNCTIONS. 


17 


RELATIONS   BETWEEN  FUNCTIONS   AND   SEGMENTS. 


11.   In  each  quadrant 

ordinate  of  C 


distance  of  0_i 


distance  of  C     ordinate  of  C 

That  is,  sin  a;  esc  ic  =  1 1 

Similarly,  tan  x  cot  a;  =  1 

sec  a?  cos  a;  =  1 


(1) 


Therefore  sin  x  is  the  reciprocal  of  esc  x,  esc  x  is  the 
reciprocal  of  sin  x ;  for  this  reason,  sin  x  and  esc  x  will  be 
called  reciprocal  functions  of  a;;  similar  statements  may  be 
made  of  the  other  functions  in  group  (1). 

In  each  quadrant, 

ordinate  of  C  _  ordinate  of  C     abscissa  of  C 
distance  of  C     abscissa  of  C     distance  of  G 


That  is, 
Similarly, 


sin  X  =  tan  x  cos  x 
tan  X  =  sec  x  sin  x 
sec  X  =  CSC  X  tan  x 
csaa;  =  cot  x  sec  x 

cot  X  =  cos  X  CSC  X 

cos  a;  =  sin  a;  cot  x 


(2) 


Solving  the  equations  in  group  (2)  for  each  function. 


sin  a; 

tana; 

cos  a; 

sec  a; 

cot  a; 

tana; 

sec  a; 
CSC  a; 

sin  a; 

cos  ay- 

esc  a; 

tana; 

CSC  a;: 


cota;  = 


cot  X      sm  X 


cos  X  = 


cot  a; 
cos  a;" 
cos  a; 
sin  a; 
sin  a; 


sec  a; 
tana; 
CSC  a; 
sec  a; 
cot  a; 


tan  X     esc  x 


(3) 


18 


FUNCTIONS   OF  ANGLES. 


Value  of  each   function. 

If  X  is  an  angle  of  the 
first  quadrant,  the  defini- 
tions give 
a 


g   sm X  =  -'^  CSC X  = 


tan  x  =  ~:   cot  x  =  - 
c  a 


sec  a; 


cos  a;  =  V 


(4) 


For  the  second  quadrant, 


a  a  b  b 

sm  ic  =    -  =  +  - ;  CSC  a;  =    -   =  4-  - 
00  a  a 

a            ci        ^         —  c         c 
tana;= = ;  cota^  = =  — 

—  c  c  a  a 

b  b  —  c          c 

sec  X  = = ;  cos  X  =  -^—  =  —  - 

—  c  c'  b  b 


(5) 


And  similarly  for  the  other  two  quadrants. 

Values  of  the  ordinate,  distance,  and  abscissa  of  any  point 
on  the  terminal  line  of  an  angle.  —  Clearing  the  equations  in 
(4)  of  fractions  gives  for  the  first  quadrant, 

a  =  b  sin  x  =  c  tan  x 

b  =  csecx  =  a(iSGx  ■  (6) 

c  =a  cot  x=b  cos  x 
The  group  in  (5)  gives  for  the  second  quadrant. 


a  =  b  sin  x  =  —  c  tan  x 
b  =  —  c  secx  =  acsGx 
c  =  —  a  cot  x  =  —  b  cos  x 
And  similarly  for  the  other  two  quadrants. 


(7) 


MNEMONIC  DEVICE. 


19 


sinjr 


coijr 


12.  Device  for  memorizing  relations.  —  The  diagram,  Fig. 
19,  with  the  rules  following,  will  be  found  useful  in  remem- 
bering    the      relations 

given   in  the  groups  of  secjr 

equations  (1)  to  (7).    In  "  ^     tan^r 

the  ligure,  the  angle  x  is 
drawn  for  the  first  quad- 
rant; but  if  the  proper 
algebraic  signs  of  a  and 
c  are  supplied,  the  dia- 
gram and  all  the  rules 
apply  to  an  angle  of 
any  quadrant. 

It  will  be  seen  that  three  cross-lines  are  drawn  through 
a  and  6,  a  and  c,  h  and  c.  Starting  with  the  right  angle 
and  going  in  the  positive  direction  of  angles,  the  cross-lines 
are  marked  sin  a;,  tana;,  seca;,  esc  a;,  cot  a;,  cosaj,  respectively. 

By  going  along  the  cross-lines,  the  relations  in  groups 
(4),  (5),  (6),  and  (7)  are  given  by  the  two  following  rules : 

Rule  I.  Any  segment  is  equal  to  the  product  of  the  two  adja- 
cents ;  thus  along  the  horizontal  cross-line,  a  =  5  sin  a;  and 
5  =  a  CSC  X. 

Rule  II.  Any  segment  or  function  is  equal  to  the  quotient  of 
the  adjacent  by  the  next ;  thus  along  the  horizontal  cross-line, 


sm  a;  =  -,  a 


CSC  a; 


-,  6  = 


a 


CSC  X 


sin  a; 


By  going  around  either  way  outside  the  triangle,  consider- 
ing only  the  functions,  equations  (2)  and  (3)  are  given  by 
two  rules : 


Rule  III.   Any  function  is  equal  to  the  product  of  the  two 
adjacents ;  thus  sin  x  =  tan  x  cos  x. 


20  FUNCTIONS  OF  ANGLES. 

EuLE  IV.  Any  function  is  equal  to  the  quotient  of  the  adja- 
cent by  the  next ;  thus 

tancc  .  cos  a; 

sm  X  = or  sm  x  = 

sec  a;  cota^ 

Considering  functions  on  the  same  cross-line,  group  (1)  is 
given  by 

KuLE  V.  Any  function  is  equal  to  the  reciprocal  of  its  oppo- 
site function ;  thus  sin  a;  esc  ic  =  1. 

ABSOLUTE   VALUES  AND 
ALGEBRAIC   SIGNS  OF  FUNCTIONS. 

13.  Definitions.  The  numerical  value  of  a  magnitude  is 
the  ratio  of  the  magnitude  to  another  magnitude  of  the 
same  kind  taken  as  a  unit.  The  absolute  value  of  a  magni- 
tude is  the  numerical  value  of  the  magnitude  considered 
without  regard  to  sign.  Thus  the  numerical  value  of 
—  5  feet  is  —  5,  and  the  absolute  value  is  simply  5. 

Absolute  values  of  functions  of  angles  in  terms  of  functions  of 
angles  between  0°  and  90°. — In  Fig.  17,  denote  the  positive 
acute  value  of  angle  BACi  by  x.  Let  the  least  positive 
value  of  angle  BAC^  =  180°  -  x,  of  angle  BAC^  =  180°  +  x, 
and  of  angle  BAC^  =  360°  —  x.  Also  make  the  distances 
ACi,  AC2,  ACs,  ACi  all  equal.  Then  the  triangles  BACiy 
B1AC2,  BiACs,  BAC^  are  all  equal. 

When  the  revolving  half-line  is  at  B  or  Bi,  the  angles 
generated  from  AB  are  all  included  in  the  expression  n  times 
90°,  where  n  is  an  even  whole  number.  And  when  the 
revolving  half-line  is  at  D  or  Di,  the  angles  generated  from 
AB  are  all  included  in  the  expression  n  times  90°,  where  n 
is  an  odd  number.  Hence  the  angles  generated  when  the 
terminal  line  is  at  ACi,  AC2,  ACs,  or  AC4  are  included  in  the 
expressions  90n  ±  x  where  n  is  even,  and  90n  ±  y  where  n 
is  odd. 


REDUCTION-  TO   FIRST  QUADRANT. 


21 


From  the  equality  of  the  various  triangles  the  ordinates 
and  abscissse  of  Ci,  Cg,  Cg,  and  C^  are  all  equal  in  absolute 
value.     Hence  the  functions  of  BACi,  BAC23  -B-^Cs,  BACi 


are  equal  in  absolute  value  to  the  same  functions  of  x ;  and 
therefore  also  equal  to  the  cof unctions  of  y.     That  is : 

I.  The  absolute  value  of  any  function  of  an  angle  dOn  ±  x 
where  n  is  even,  is  the  same  as  the  absolute  value  of  that  func- 
tion of  X. 

II.  TJie  absolute  value  of  any  function  of  an  angle  90  /?  ±  / 
where  n  is  odd  is  the  same  as  the  absolute  value  of  that  co- 
function  of  ^. 

Expressed  in  a  formula,  /  standing  for  any  function : 

/(90  n  ±  x)=  ±fx  when  n  is  even  =  ±  cofx  when  n  is  odd. 

For  example  :  Any  function  of  150°,  210°,  330°,  390°,  510°, 
—  30°,  —  150°,  etc.,  is  equal  in  absolute  value  to  the  same 
function  of  30°  or  the  co-named  function  of  60°. 


22  FUNCTIONS  OF  "ANGLES. 

The  algebraic  signs  of  the  functions  may  be  remembered  by 
this  device : 

Write  the  functions  in  the  order 

sin    tan    sec    esc    cot    cos 

2        3        4        2        3       4 

The  functions  with  positive  values  are  : 

In  Quadrant  I.        All  of  them. 

In  Quadrant  II.  sin     esc 

In  Quadrant  III.       tan    cot 

In  Quadrant  IV.         sec     cos 

In  all  other  cases  the  functions  are  negative  in  value. 

The  method  of  obtaining  the  value  of  a  function  of  any  angle 
in  terms  of  an  angle  between  0°  and  90°  is  given  in  these 
two  rules : 

Rule  I.  Divide  the  angle  by  90°.  If  the  quotient  is  even, 
take  the  same  function  of  the  remainder.  If  the  quotient  is  odd, 
take  the  cofunction  of  the  remainder. 

Rule  II.    Prefix  the  proper  algebraic  sign. 

Thus  sin  172°=  cos  82°.  esc  274°  9'=- sec  4°  9' 
tan  305°  =  -  cot  35°.  cot  329°  51 '  =  -  tan  59°  51 ' 
sec  595°  =  -  sec  55°.     cos  935°  22'  =  -  cos  35°  22' 

Notice  that  the  quadrant  is  indicated  by  1  +  the  quotient. 

14.  Functions  of  180°  ±  x,  of  90°  +  x,  and  of  -  jr.  —  Con- 
sider any  function,  say  the  sine,  of  the  two  angles  180°  —  x 
and  X.  It  has  been  shown  that  sin  (180°  —  x)  and  sin  x  are 
equal  in  absolute  value.    Consider  now  their  algebraic  signs. 

If  x  is  of  the  first  quadrant,  180°  —  a;  is  of  the  second 
quadrant ;  then  sin  x  and  sin  (180°  —  x)  are  both  +. 


REDUCTION  TO  FIRST  QUADRANT. 


23 


If  X  is  of  the  second  quadrant,  180°  —  x  is  of  the  first 
quadrant ;  then  sin  x  and  sin  (180°  —  x)  are  both  -|-. 

If  x  is  of  the  third 
quadrant,  180°  -  a;  is  of  I> 

the  fourth  quadrant ; 
then  sinic  and  sin 
(180° -a;)  are  both  -. 

If  X  is  of  the  fourth 
quadrant,  180°  —  x  is 
of  the  third  quadrant; 
then  sin  a;  and  sin 
(180° -a;)  are  both  -. 

Hence  the  equation 
sin  (180°  —  a;)  =  sin  x  is 
true  for  all  values  of  x. 

In  a  similar  manner,  it  may  be  shown  that  the  equation 
sin  (90°  +  a;)  =  cos  x  is  true  for  all  values  of  x.  Also  that 
sin  (180°  +  a;)  =  sin  (—  a;)  =  —  sin  a;  are  true  for  all  values 
of  X. 

In  a  similar  manner,  other  functions  of  these  angles  may- 
be considered.    The  results  are  shown  in  the  following  table  : 

sin  (90°  ■\-x)=  cos  x. 
tan  (90°  +  a;)  =  —  cot  x. 
sec  (90°  -\-x)  =  —  CSC  x, 
esc  (90°  +  a;)  =  sec  x. 
cot  (90°  +  a;)  =  - tan  a;, 
cos  (90°  +  a;)  =  —  sin  x. 

sin(— a;)=  — sina;. 
tan  (—  a;)  =  —  tan  x. 
sec  (—a;)  =  sec  x. 
csc(— a;)  =  — CSC  X. 
cot  (—a;)  =  — cot  x. 
cos(— a;)=     cos  a;. 


sin  (180° 

-x)^ 

sin  a?. 

tan  (180° 

-x)  =  - 

-  tanx. 

sec  (180°  - 

-x)  =  - 

-  sec  x. 

CSC  (180°  - 

-x)  = 

CSC  X. 

cot  (180°  -x)  =  -  cot  X. 
cos  (180°  —  x)  =  —  cos  X. 

sin  (180°  -\-x)  =  —  sin  x. 
tan  (180°  + a;)  =  tana;, 
sec  (180°  +  a;)  =  —  sec  x. 
esc  (180°  +  a;)  =  —  esc  x. 
cot  (180°  +  x)=  cot  X. 
cos  (180°  -\-x)  =  —  cos  X. 


24 


FUNCTIONS  OF  ANGLES. 


THE  CURVE  OF  EACH  FUNCTION. 

15.  If  a  point  moves  so  that  its  distance  remains  the 
same,  the  ordinate  of  the  moving  point  will  be  proportional 
to  the  sine  of  the  varying  angle;  and  the  abscissa  of  the 
moving  point  will  be  proportional  to  the  cosine  of  the  angle ; 
for  the  sine  of  the  angle  equals  the  ordinate  divided  by  the 
distance;  and  the  cosine  of  the  angle  equals  the  abscissa 
divided  by  the  distance ;  and  therefore  if  the  distance  re- 
mains the  same,  the  sine  varies  as  the  ordinate ;  and  the 
cosine  varies  as  the  abscissa. 

Suppose,  now,  a  second  point  to  move  in  the  plane  so 
that  its  abscissa  is  proportional  to  the  angle,  and  its  ordi- 
nate is  proportional  to  the  sine  of  that  angle.  As  the  angle 
changes  from  0°  to  360°,  this  second  point  will  trace  out  the 


Fig.  20.  —  Curve  of  sines. 


Curve  of  cosines. 


curve  shown  in  Fig.  20,  called  a  simple  harmonic  curve,  or 
curve  of  sines.  Or,  if  the  ordinate  of  the  second  point  was 
proportional  to  the  cosine  of  the  angle,  the  curve  is  called 
the  curve  of  cosines. 

If  the  first  point  moves  in  a  line  at  right  angles  to  the 
initial  line,  the  ordinate  of  this  point  will  be  proportional 
to  the  tangent  of  the  angle,  and  the  distance  of  this  moving 
point  will  be  proportional  to  the  secant  of  the  angle ;  now 
if  the  abscissa  of  the  second  moving  point  was  proportional 
to  the  angle  and  the  ordinate  proportional  to  the  tangent 
of  the  angle,  the  curve  traced  out  is  called  the  curve  of 
tangents  (Fig.  21) ;  or,  if  the  ordinate  of  the  second  point 


CURVES  OF  FUNCTIONS. 


26 


\               / 

\           / 

V                 / 

\          / 
\        / 

\    / 

\     / 
\  / 

A 

/  \ 

/      \ 

/          \ 

IW 

\                                1 

\                            / 

\                          / 

\                     / 

•  \                  / 

\              / 

\         / 

\      / 

/\ 
/         \ 

/                \ 
/                          \ 
/                                \ 

36o' 

90- 

\         / 

/          \ 

/                 \ 
/                     \ 
/                        \ 

/                            \ 
1                              \ 

no' 

\                              / 
\                        / 

\      / 

/\ 

/         \ 

/              \ 

/                  \ 

/                       \ 

/                          \ 

/                             \ 

/                                \ 

Fig.  21. —Curve  of  tangents.   Curve  of  cotangents. 


\              / 

/ 

\ 

\               / 

/ 

\ 

\              / 
\          / 

\           / 

/ 
/ 

\ 

\.         / 

/ 

\ 

'y/ 

/ 
/ 

\ 

^         ^^ 

^ 

c" 

.^- 

iSo* 

»70*                   .,6o- 

/^ 

^Vv^'' 

^N 

/ 

/     \ 
/        \ 
/           \ 

\ 

\ 

\ 

/ 

/           \ 

\ 

/ 

/             \ 
/                 \ 

/              \ 

\ 
\ 

\ 

Fig.  22.  — Curve  of  secants. 


Curve  of  cosecants. 


26  FUNCTIONS  OF  ANGLES. 

was  proportional  to  the  secant  of  the  angle,  the  curve  is 
called  the  curve  of  secants  (Fig.  22). 

In  a  similar  manner,  the  first  point  may  be  assumed  to 
move  parallel  to  the  initial  line,  and  the  second  point  may 
be  made  to  trace  the  curve  of  cosecants  (Fig.  22),  or  the  curve 
of  cotangents  (Fig.  21). 

LIMITING  VALUES   OF  FUNCTIONS. 

16.  Since  from  geometry  the  hypotenuse  of  a  right  tri- 
angle can  never  be  less  than  either  side,  the  distance  of  a 
point  can  never  be  less  than  either  the  abscissa  or  ordinate 
of  the  point.  Therefore  the  absolute  value  of  the  sine  or 
of  the  cosine  of  any  angle  is  never  more  than  1,  and  the 
absolute  value  of  the  secant  or  of  the  cosecant  of  any  angle 
is  never  less  than  1.  The  absolute  value  of  the  tangent 
or  of  the  cotangent  of  angles  is  without  limit. 

THE    ANGLES    EACH    LESS    THAN    360°    CORRESPONDING 
TO  A  GIVEN  VALUE   OF  A  FUNCTION.* 

17.  From  what  has  been  said,  it  follows  that  there  is 
but  one  value  of  any  function  of  a  given  angle ;  but  for  a 
given  numerical  value  of  any  function  of  the  angle,  there 
will  be  0,  1,  or  2  values  of  the  angle  between  0°  and  360°. 

1.   If  sin  X  or  cos  x  is  more  than  1  or  less  than  —  1,  or  if 

sec  X  or  esc  x  is  between  1  and  —l,x  will  be  impossible. 

2.  If  sin  a?  =  H- 1  or  esc  x=  -rl,  x=   90°. 

If  sin  a;  =  —  1  or  CSC  a?  =  —  1,  x  =  270°. 

If  sec  a;  =+  1  or  cos  x  =  -|-  1,  x  =     0°. 

If  sec  a;  =  —  1  or  cos  a;  =  —  1,  a;  =  180°. 

3.  In  all  other  cases,  two  values  of  x  between  0°  and  360° 
will  be  obtained. 

*  This  section  can  be  more  easily  followed  if  the  diagrams,  Figs.  20, 
21,  and  22,  are  referred  to. 


TO  FIND  AN  ANGLE   FROM  A  FUNCTION.  27 

Suppose  that  sin  jr  =  iV  (or  esc  x  =  N),  and  that  a^i  is 
the  corresponding  acute  value  of  x.  Then  x  also  equals 
180°  -  X,. 

If  the  given  equation  is  sin  x=  —  N  (or  esc  x=  —  N) 
and  sin  Xi  =  N  (or  esc  x^  =  N),  where  x^  is  acute,  then 

X  =  180°  +  a^i  or  a;  =  360°  -  Xi. 

If  tan  x=  N (or  cot  x  =  N),  and  Xi  is  the  acute  value  of  x, 
another  value  of  x  is  180°  +  Xi. 

If  the  given  equation  is  tan  x  =  —  N  (or  cot  x  =  —  N) 
and  tan  Xi  =  N  (or  cot  Xi  =  N)  where  Xi  is  acute,  then 

x  =  180°-Xi   or   a;  =  360°-aj. 

If  sec  jr  =  -^  (or  cos  x  =  N)  and  iCi  is  the  acute  value  of  x, 
another  value  of  x  is  360°  —  x^. 

If  the  given  eqviation  is  sec  a;  =  —  ^  (or  cos  x  =  —  N) 
and  sec Xi=  N  (or  cos x^  =  N)  where  Xi  is  acute,  then 

a;  =  180°-a;i   or   a;=180°  +  a:i. 

Angle  less  than  180°.  —  If  the  problem  of  obtaining  the 
values  of  an  angle  from  a  given  function  of  the  angle  is 
still  further  limited  by  the  condition  that  the  angle  shall 
be  less  than  180°  (say  an  angle  of  a  triangle),  then  : 

(1)  If  the  given  function  is  a  sine  or  a  cosecant  with  a 
negative  value,  no  value  of  the  angle  could  be  obtained ; 
because  all  angles  between  0°  and  180°  have  a  positive  sine 
and  a  positive  cosecant. 

(2)  If  the  given  function  is  a  sine  or  a  cosecant  with  a 
positive  value,  and  the  given  value  of  the  function  was  pos- 
sible, there  will  be  two  supplementary  values  of  the  angle ; 
because  if  Xi  is  the  acute  value,  180°  —  Xi  will  be  the  obtuse 
value.  If  x^  happens  to  be  90°,  it  may  be  said  that  the 
two  values  are  equal,  or  there  is  but  one  value. 

(3)  If  the  given  function  of  the  required  angle  is  any 
other  function  than  the  sine  or  cosecant,  there  can  be  but 


28  FUNCTIONS   OF  ANGLES. 

one  value  of  the  angle  less  than  180° ;  acute  if  the  value  of 
the  function  is  positive;  obtuse  if  the  value  of  the  func- 
tion is  negative. 

If  it  is  known  that  a  required  angle  was  not  greater  than 
90°,  a  negative  value  of  any  'function  of  the  angle  would 
indicate  that  no  value  of  the  angle  was  possible. 

FUNCTIONS   OF  ^x  AND   2  X. 

18.  Functions  of  i  jr.  —  Bisect 
angle  BAC,  and  draw  BJE  parallel 
to  AD.  Denote  angle  BAC  by  x. 
Then  angle  E  =  angle  ABE  =  an- 
gle BAD  =  ix.     Also  AE  =  AB. 

BC^  =  AC^-AB'.     Divide   by 

BC. 

1  =  CSC-  X  —  cot^  X.  (8) 

AB^EA^EC^AC     AB 
BD     BD     BC     BC     BC' 

cot  i  cc  =  CSC  X  -f-  cot  X  = 

smx 

EB     EC       ,   BD     BC       .  EB  ^  BD      BC 

=  — -  and  = ;  whence  — —  x •  =  — -^• 

AD     AC  EA     EC'  AE     AD     AC 

2  cos  i  a;  sin  i  a;  =  sin  x.  (10) 

The  quotient  of  (8)  and  (9)  is 

tan  4- aj  =  esc  a;  —  cot  a;  = -^ (11) 

^  sma;  ^ 

The  product  of  (10)  and  (11)  is 

2  sin^  i  a;  =  1  —  cos  x.  (12) 

The  product  of  (9)  and  (10)  is 

2cos2|-a;  =  l-f  cosa;.  (13) 

The  half  difference  of  (13)  and  (12)  is 

cos^ ^x  —  sin^ i a;  =  cos  x.  (14) 


FUNCTIONS  OF  PARTICULAR  ANGLES.       29 

The  reciprocal  of  (10)  is 

^  sec  ^  a;  CSC  ^  it-  =  esc  x.  (15) 

Functions  of  2  x.  —  Denote  now  the   angle   BAD  by  x. 
Then  from  (10),  (14),  and  (15) : 

sin  2  a;  =  2  sin  x  cos  x.  (16) 

cos  2x  =  cos'^ X  —  sin^ x.  (17) 

CSC  2  a;  =  "I"  sec  a;  esc  x.  (18) 

The  other  functions  can  then  be  found  in  terms  of  these. 

FUNCTIONS  OF  PARTICULAR  ANGLES. 

19.  Functions  of  45°,  22°  30',  etc.  — Let  angle  ^  =  45°. 
Denote  the  length  of  AB  by  c ;  then  BC  =  c 
and  AC  =  V^^T^^  =  1-4142  c.     Hence 

tan  45°=  cot  45°  =  1. 

sec  45°  =  esc  45°  =  1.4142. 

Also,        sin  45°  =  cos  45°  =  .7071.  AZ ^b 

Fig.  24. 
From  these  values  of  the  functions  of  45°, 

the  values  of  the  functions  of  22°  30'  can  be  found  by  means 
of  formulae  (9)  to  (13).  Then  from  the  values  of  the  func- 
tions of  22° 30',  the  functions  of  11°  15'  can  be  found;  and 
so  on. 

Functions  of  60°,  30°,  15°,  etc.  — The  tri-  C 

angle  having  angles  of  90°,  60°,  and  30°  is 
one-half  of  an  equilateral  triangle. 

Take  BC=a;  then  AC  =2  a  and 


AB  =\^4.a'-a^  =  1.7321  a.  ^^^,; 

Fig.  25. 
Hence  the  functions  of  60°  and  30°  can 

be  computed.     And,  therefore,  the  functions  of  15°,  7°  30', 

etc.,  may  be  computed. 


30 


FUNCTIONS   OF  ANGLES. 


Functions  of  72°,  36°,  18°,  etc.  — Let  ^C=  CD,  and  DE 

bisect  angle  D,  CB  bisect  angle  (7;  take  the  value  of  angle  A 
to  be  72°.     All  the  other  angles  on  the  fig- 
ure can  then  be  found.     From  the  equality 
of  angles,  AD  =  ED  =  EC ;    also  the  tri- 
angle ACD  is  similar  to  the  triangle  ADE. 

Denote  the  length  of  ^C  by  h,  and  the 
length  of  ^5  =6  cos  72°  by  hx.  Then 
AE=b-'2hx. 

From  the  similar  triangles  ACD  and 
ADE, 

AG^^AD    ^^      b    ^     2bx 
AD     AE    °^    2bx     b-2bx 

Solving :   a?  =  cos  72°  =  sin  18°  =  i  ( V5  -  1)  =  .3090. 

Denote  the  length  of  BC  by  by ;  then 

BC  =  AC  -  AB'  or  by  =  6^  _  52^ 

That  is,    y'=l-x^  01  sin^  72°  =  cos^  18°  =  1  -  .30901 

Hence  the  other  functions  of  72°  and  18°  may  be  found. 
Also  the  functions  of  36°,  9°,  4°  30',  etc.,  may  be  found. 

Exercises. —  For  all  values  of  x,  prove  that 

sin^  X  -\-  cos^  ic  =  1, 
sec^aj  —  tannic  =  1, 
csc^  X  —  cot^  X  =  1, 

and  explain  how  all  the  functions  of  an  angle  may  be  found 
when  the  value  of  only  one  function  is  known. 


CHAPTER   II. 

EXPLANATION   OF  THE  TABLES. 
I.   NATURAL  VALUES   OF  FUNCTIONS. 

20.  The  values  given  for  the  functions  of  angles  in  The 
Table  of  Natural  Values  of  Functions,  page  i  to  page  vii, 
are  approximate  values  to  four  significant  figures  at  least. 
The  last  figure  given  in  each  value  is  the  nearest  figure  in 
that  place.  Thus  the  value  of  sin  32°,  which  is  not  less 
than  .52985  and  not  greater  than  .52995,  is  given  on  page  v, 
line  25,  as  .5299. 

Any  value  printed  without  a  decimal  point  will  have  one 
or  more  figures  missing.  The  decimal  point  and  the  missing 
figures  will  be  found  to  the  left  and  slightly  above  the  fig- 
ures printed  for  the  value.  Thus  page  v,  line  29,  sin  32°  40' 
is  .5398,  the  398  being  printed  in  the  proper  place  and  .5 
being  found  higher  and  to  the  left. 

tan  32°  40'  =  .6412,  sec  32°  40'  =  1.1879,  etc. 

Given  an  angle  to  find  any  function  of  the  angle — Angles 
between  0°  and  45°  will  be  found  in  the  column  marked  x 
on  the  left  of  the  page,  and  angles  between  45°  and  90°  in 
the  column  marked^  on  the  right.  When  the  angle  is  found 
on  the  left,  the  names  of  the  functions  are  given  at  the  top 
of  the  columns,  x  being  added  to  the  name  of  each  function. 
When  the  angle  is  found  on  the  right,  the  names  of  the 
functions  (marked/)  are  given  at  the  bottom  of  the  page; 
thus : 

31 


32         EXPLANATION  OF  THE  TABLES. 

Page  V,     line  39  sin  34°  20'  =    .5640 

Pagevi,   line    9  tan  53°  40' =  1.3597 

Page  iv,   line    2  sec  21°  10'  =  1.0723 

Page  vii,  line  16  esc  45°  30'  =  1.4020 

Pageiv,   line  26  cot  64°  50'=    .4699 

Page  iv,   line    7  cos  22°        =    .9272 

To  find  function  of  angles  which  are  not  multiples  of  10'.  — 

When  we  desire  the  function  of  an  angle,  whose  value  is 
between  two  angles  of  the  table,  a  process  called  interpolation 
is  used. 

By  reference  to  any  part  of  the  table,  it  will  be  seen  that 
small  changes  in  the  angle  produce  nearly  proportional  changes 
in  the  values  of  any  function  of  the  angle.  A  four-place  table 
usually  gives  the  values  of  the  functions  of  angles  for  every 
10'.  By  assuming  that  the  change  in  any  function  is  pro- 
portional to  small  changes  in  the  angle,  the  value  of  the 
function  of  an  angle  to  a  single  minute  can  in  general  be 
found  to  four  significant  figures  without  appreciable  error. 
Conversely,  if  the  value  of  the  function  is  correct  to  four 
significant  figures,  the  value  of  the  angle  can  in  general  be 
found  correct  to  a  single  minute. 

A  five-place  table  giving  the  values  of  the  functions  for 
each  minute  can  be  used  to  interpolate  to  10";  a  six-place 
table  for  every  10"  can  be  interpolated  to  1" ;  a  seven-place 
table  for  every  second  can  be  interpolated  to  .1",  and 
so  on. 

With  every  increase  in  the  number  of  significant  figures, 
the  degree  of  precision  is  increased  and  also  the  labor  of 
using  the  tables.  If  the  data  of  a  computation  contain 
angles  measured  to  a  single  minute,  four-place  tables  can 
be  used  in  the  computation;  if  the  angles  are  measured 
to  one-tenth  second,  seven-place  tables  can  be  used;  and 
so  on. 


INTERPOLATION.  33 

Let  /stand  for  any  function.  Then  if  h  is  a  small  addition 
to  angle  x,  and  z  is  the  change  in  any  function  of  x  due  to  a 
change  k  (10')  in  the  angle,  it  is  assumed 

f(x  +  h)-fx^h 
z  k 

or  f{x^-h)=fx  +  h, 

K 

from  which  the  value  of  a  function  of  an  angle  x  +  h  can  be 
obtained  by  substitution.     Thus : 

sin  34°  23'  =  sin  34°  20'  +  .3  (sin 34°  30'  -  sin  34°  20') 

=  .5640  -f  .3  X  .0024  =  .5647. 
tan 53°  44'  =  tan 53°  40'  +  .4(tan53°  50'  -  tan 53°  40') 

=  1.3597  +  .4  X  .0083  =  1.3630. 
sec  21°  11'  =  sec  21°  10'  +  .1  (sec  21°  20'  -  sec  21°  10') 

=  1.0723  +  .1  X  .0013  =  1.0724. 
esc  45°  38'  =  CSC  45°  30'  +  .8  (esc  45°  40'  -  esc  45°  30') 

=  1.4020  +  .8  X  (-  .0040)  =  1.3988. 
cot  64°  55'  =  cot  64°  50'  +  .5  (cot  65°  -  cot  64°  50') 

=  .4699  +  .5  (-  .0036)  =  .4681. 
cos  22°  7'  =  cos  22°  +  .7  (cos  22°  10'  -  cos  22°) 

=  .9272  +  .7  (-  .0011)  =  .9264. 

It  should  be  noted  that  all  of  the  co-named  functions  of 
acute  angles  decrease  as  the  angle  increases.  Hence  when 
the  tabulated  value  of  a  co-named  function  of  an  acute 
angle  is  corrected  for  an  addition  to  the  angle,  the  cor- 
rection will  be  negative. 

If  functions  of  angles  greater  that  90°  are  desired,  they 
are  obtained  by  applying  the  rules  on  page  22 ;  thus  (Quad- 
rant II), 


34         EXPLANATION  OF  THE  TABLES. 

sin  155°  12 '  =  cos  65°  12 '  =  .4195. 
tan  113°  47'  =  -  cot  23°  47'  =  -  2.2691. 
sec  167°  9'  =  - CSC  77°  9' =  -1.0257. 
CSC  123°  42'  =  sec  33°  42'  =  1.2020. 
cot  147°  56'  =  -  tan  57°  56'  =  -  1.5962. 
cos  134°  35'  =  -  sin  44°  35'  =  -  .7019. 
And  in  a  similar  manner  for  angles  of  Quadrants  III 
and  IV. 

Given  the  value  of  a  function  of  the  angle,  to  find  the  value  of 
that  angle.  —  The  table  is  arranged  to  give  the  value  of  the 
function  when  the  angle  is  known.  The  reverse  operation 
of  obtaining  the  angle  when  the  function  is  known  is  some- 
what more  inconvenient.  Assume  that  we  know  the  sine 
of  the  required  angle ;  the  given  value  may  be  sought  either 
in  the  column  marked  sin  x  at  the  top  of  the  page  or  in  the 
column  marked  sin/  at  the  bottom  of  the  page.  If  the 
number  is  found  in  the  column  marked  sinjr,  and  it  is 
known  that  the  angle  is  acute,  the  value  of  the  angle  will 
be  found  in  the  column  marked  jr  on  the  left  of  the  page. 
If  the  given  value  of  the  sine  is  found  in  the  column  marked 
sin/,  the  value  of  the  acute  angle  will  be  found  in  the  col- 
umn marked  /.  Any  other  function  will  be  found  in  a 
similar  manner;  thus: 

Page  V,  line  27.  If  sin^  =  .8450,  A  (acute)  =  57°  40'. 
Page  V,  line  4.  If  tan  A  =  1.8418,  A  (acute)  =  61°  30'. 
Page  V,  line  41.  If  sec  A  =  1.2158,  A  (acute)  =  34°  40'. 
Page  iv,  line  33.  If  esc  A  =  2.2543,  A  (acute)  =  26°  20'. 
Page  V,  line  34.  If  cot  ^  =  .6619,  A  (acute)  =  56°  30'. 
Page  vi,  line  27.  If  cos  A  =  .7735,  A  (acute)  =  39°  20'. 
If  the  values  of  the  functions  of  45°  are  memorized 
(sin  45°  =  .7,  tan  45°  =  1,  sec  45°  =  1.4), 


INTERPOLATION  TO  FIND  ANGLE.  35 

it  will  be  known  whether  the  given  value  of  the  functions 
should  be  sought  in  the  columns  on  the  right  of  the  page  or 
on  the  left.  Thus  if  sin  A  =  .9112,  the  value  of  A  must  be 
more  than  45°,  because  the  sine  of  an  acute  angle  increases 
as  the  angle  increases,  and  the  sine  of  45°  is  .7.  Since  the 
angle  is  more  than  45°,  the  given  value  is  sought  in  the 
column  marked  sin  y. 

As  before,  let  /  stand  for  any  function,  h  a  small  addition 
to  angle  x,  and  z  a  change  in  any  function  due  to  a  change 
Jc  (10')  in  the  angle.  Then,  since  small  changes  in  the  angle 
are  nearly  proportional  to  the  changes  in  any  function, 

k  z 

,  f(x-\-  Ji)  —  fx  - 
x-\-h  =  x-{-''^         ' — —k. 
z 

If  sin  A  =  .6479,  the  acute  value  of  A 

=  40°  20'  +  •^^'^9  -  •^^'^'^  10'  =  40°  23'. 
.6494 -.6472 

If  tan  A  =  3.7539,  the  acute  value  of  A 

—  7K°  _!_  3.7539  —  3.7321  -tm  _  7^0  Kf 

If  sec  A  =  1.2190,  the  acute  value  of  A 

^  3^0  50,  ^  1.2190-1.2183  ^0,  ^  3^0  53, 
^1.2208-1.2183 
If  esc  A  =  3.9984,  the  acute  value  of  A 

=  14°  20'  +  ^-^^^^  -  ^-^^^^  10'  =  14°  29'. 
3.9939  -  4.0394 

If  cot  A  =  2.2062,  the  acute  value  of  A 

=  24°  20'  +  ^-^Q^^  -  2.2113  ^Q,  ^  24°  23'. 
2.1943-2.2113 

If  cos  A  =  .5654,  the  acute  value  of  A 

=  55°  30'  -f  '^^^^  ~  ^^^^  10'  =  55°  34'. 
^  .5640  -  5664 


36  EXPLANATION   OF  THE   TABLES. 

If  both  values  of  the  angle  which  are  less  than  360°  are 
desired,  any  value,  not  acute,  can  be  obtained  by  a  reversal 
of  the  rules  on  page  22 ;  thus. 

If  sin  A  =    .6479,  A  =  40°  23'  or  139°  37'. 

If  tan^  =  3.7539,  A  =  75°    5'  or  255°   5'. 

If  sec  A  =  1.2190,  A  =  34°  53'  or  325°    7'. 

If  CSC  A  =  3.9984,  A  =  14°  29'  or  165°  31'. 

If  cot  A  =  2.2062,  A  =  24°  23'  or  204°  23'. 

If  cos  A  =    .5654,  A  =  55°  34'  or  304°  26'.       . 

Also  when  the  given  function  is  negative : 

If  sin  ^  =  -    .6479,  A  =  220°  23'  or  319°  37'. 

If  tan  ^  =  -  3.7539,  A  =  104°  55'  or  284°  55'. 

If  sec  ^  =  -  1.2190,  A  =  145°   7'  or  214°  53'. 

If  CSC  ^  =  -  3.9984,  A  =  194°  29'  or  345°  31'. 

If  cot  ^  =  -  2.2062,  A  =  155°  37'  or  335°  37'. 

If  COS  ^  =  -    .5654,  A  =  124°  26'  or  235°  34'. 

It  should  be  noted  that  if  the  given  value  of  the  sine  or 
cosine  of  an  angle  is  greater  than  + 1  or  less  than  —  1,  no 
value  of  the  angle  can  be  obtained ;  and  if  the  given  value 
of  the  secant  or  cosecant  of  an  angle  is  between  +1  and 
—  1,  no  value  of  the  angle  can  be  obtained. 

Exercises.  —  Copy  one  value  in  any  row  of  the  following 
table.     Then  find  each  of  the  remaining  values  in  the  row. 

sin      0°43'  sin  179°  17'  .0125  cos   89°  17'  cos  270°  43' 

sin    10°18'  sin  169° 42'  .1788  cos   79° 42'  cos280°18' 

sin  200° 41'  sin  339°  19'  -.3532  cos  110° 41'  cos  249°  19' 

sin  210°   6'  sin  329°  54'  -.5015  cos  120°   6'  cos  239°  54' 

sin    40°  23'  sin  139°  37'  .6479  cos   49°  37'  cos  310° 23' 


LOGARITHMS. 


3T 


tan   84°  15'  tan  264°  15' 

tan   74°  18'  tan  254°  18' 

tan  157°  13'  tan  337°  13' 

tan  141°  33'  tan  321°  33' 

tan  135°  19'  tan  315°  19' 

sec   22°  34'  sec  337°  26' 

sec   63°  12'  sec  296°  48' 

sec   99°  11'  sec  260°  49' 

sec  128°  43'  sec  231°  17' 


9.9310  cot     5°  45'  cot  185°  45' 

3.5576  cot   15°  42'  cot  195°  42' 

-.4200  cot  112° 47'  cot  292° 47' 

-.7940  cot  128° 27'  cot  308° 27' 

-.9890  cot  134° 41'  cot  314° 41' 

1.0829  CSC    67°  26'  esc  112°  34' 

2.2179  CSC   26° 48'  esc  153°  12' 

-6.2659  CSC  189°  11'  esc  350° 49' 

-1.5988  csc218°43'  csc321°17' 


sec  107°  47'    sec  252°  13'    -3.2742    esc  197° 47'    esc  342°  13' 


II.     LOGARITHMS  OF  NUMBERS. 

21.  Definition.  —  Logarithms  are  a  series  of  numbers  sever- 
ally assigned  to  the  ordinary  numbers,  and  so  arranged  that 

the  multiplication  and  division 
of  ordinary  numbers  is  accom- 
plished by  means  of  the  addition 
or  subtraction  of  their  loga- 
rithms. 

Many  systems  of  logarithms 
are  possible ;  but  in  the  simplest 
systems  the  ordinary  numbers 
are  powers  of  a  number  called  a 
base,  and  the  logarithms  are  the 
corresponding  exponents. 

Thus,  if  the  base  is  2,  the  num- 
bers are  powers  of  2,  and  the  cor- 
responding exponents  are  called 
logarithms  to  the  base  2.  To 
multiply  64  by  16,  we  add  the 
logarithms  of  64  and  16  (6  and  4), 
producing  10 ;  and  10  is  the  logarithm  of  1024,  the  product. 


Numbers 

Logarithms 

OR 

OB 

Powers. 

Exponents. 

1 

0 

2 

1 

4 

2 

8 

3 

16 

4 

32 

5 

64 

6 

128 

7 

256 

8 

512 

9 

1024 

10 

38         EXPLANATION  OF  THE  TABLES. 

To  divide  512  by  32,  subtract  their  logarithms  (9  and  5), 
giving  4 ;  and  4  is  the  logarithm  of  16,  the  desired  quotient. 

To  raise  4  to  the  fifth  power,  multiply  the  logarithm  of  4 
(2)  by  the  index  of  the  power  (5),  producing  10,  the  loga- 
rithm of  1024,  the  required  power. 

To  obtain  the  cube  root  of  512,  divide  the  logarithm  of 
512  (9)  by  the  index  of  the  root  (3),  giving  3,  the  logarithm 
of  8,  the  desired  root. 

Apply  to  logarithms  the  rules  of  algebra  relating  to 
exponents.     Then  we  have 

Log  of  product  =  sum  of  logs  of  factors. 

Log  of  quotient  =  log  of  dividend  -  log  of  divisor. 

Log  of  power  of  a  number  =  log  of  number  x  index  of  power. 

Log  of  root  of  a  number  =  log  of  number  ^  index  of  root. 

CHARACTERISTIC  AND   MANTISSA. 

If  we  think  of  integers  as  being  arranged  with  zeros 
expressed  decimally  (thus,  7  =  7.0000),  all  logarithms  may 
be  said  to  consist  of  two  parts,  —  an  integral  part  and  a 
fractional  part. 

The  integral  part  of  a  logarithm  is  called  its  characteristic. 
The  fractional  part  is  called  its  mantissa.  Thus,  if  2.1347 
is  a  logarithm,  2  is  its  characteristic,  and  .1347  is  its 
mantissa. 

The  characteristic  may  be  either  positive  or  negative,  but 
the  logarithm  is  so  arranged  that  the  mantissa  is  always 
positive.  Thus,  if  the  logarithm  was  —  2.4178,  the  number 
3  would  be  added  and  subtracted  thus,  3  —  2.4178  —  3  = 
.5822  —  3.  In  this  logarithm,  —  3  is  the  characteristic,  and 
.5822  is  the  mantissa. 

Exercises.  —  Find  the  characteristic  and  mantissa  of  the 
following  logarithms :    4|-,   —  If,  -|,   —  i^-,   —  f . 


CHARACTERISTIC  AND  MANTISSA.  39 

DENARY  LOGARITHMS. 

Although  a  practical  table  of  logarithms  could  be  con- 
structed by  taking  any  positive  number  as  the  base  of  the 
system,  the  common  system  of  logarithms  has  10  for  its 


The  system  whose  base  is  10  has  two  principal  advan- 
tages : 

(1)  The  mantissa  of  a  logarithm  depends  only  on  the 
figures  of  the  number  and  their  order;  or,  to  express  it 
another  way,  the  mantissa  of  the  logarithm  of  a  number 
is  independent  of  the  position  of  the  decimal  point  in  the 
number.  For  example,  the  numbers  4.57  and  4570  have 
the  same  mantissa  in  their  logarithms. 

(2)  The  characteristic  of  a  logarithm  depends  only  on  the 
position  from  the  units  place  of  the  first  significant  figure 
of  the  number ;  or,  to  express  it  another  way,  the  charac- 
teristic of  the  logarithm  of  a  number  is  independent  of  the 
figures  in  the  number  and  of  the  order  of  the  figures.  For 
example,  the  numbers  45.7  and  23.8  have  the  same  charac- 
teristic in  their  logarithms. 

To  prove  the  first  of  these  principles,  say  the  logarithm  of 
4.57  is  0.6599,  and  that  the  logarithm  of  A  is  a.     Then 

10«-^^=       4.57.  10-     =A. 

But      W      =1000  103     =1000. 

Hence  lO^-^^^  =  4570.  10" + ^  =  looo  A. 

It  will  be  noticed  that  the  decimal  point  of  a  number  is 
moved  to  the  right,  when  the  number  is  multiplied  by  an 
integral  power  of  10.  Therefore  the  logarithm  of  the  prod- 
uct will  be  increased  by  this  integer  (the  exponent  of  the 
multiplier).  But  adding  an  integer  to  a  logarithm  does  not 
affect  the  fractional  part  of  the  logarithm.  Hence  the  man- 
tissa of  the  logarithm  is  unchanged. 


40 


EXPLANATION  OF  THE  TABLES. 


In  a  similar  manner,  it  may  be  shown  that  a  motion  of 
the  decimal  point  to  the  left  will  leave  the  mantissa  of  its 
logarithm  unchanged. 

To  prove  the  second  of  these  principles  consider  the  follow- 
ing table  of  logarithms,  the  base  of  the  system  being 
10: 

From  an  inspection  of  the  table  it  will  be  seen  that  all 

numbers  between  1000  and 
10,000  (numbers  with  the 
first  significant  figure 
three  places  to  the  left  of 
the  units  place)  will  have 
logarithms  between  3  and 
4,  and  hence  3  +  a  proper 
fraction;  and  in  general 
that  numbers  with  the 
first  significant  figure  n 
places  to  the  left  of  the 
units  place  will  have  their 
logarithms  between  n  and 
n  +  1,  and  hence  ?i  +  a 
proper  fraction.  Also  that  all  numbers  between  .0001  and 
.001  (numbers  with  the  first  significant  figure  four  places  to 
the  right  of  the  units  place)  will  have  their  logarithms 
between  —  4  and  —  3,  and  hence  —  4  +  a  proper  fraction ; 
and  in  general  that  numbers  with  the  first  significant  figure 
n  places  to  the  right  of  the  units  place  will  have  their  loga- 
rithms between  —n  and  —  (?i  — 1),  and  hence  —  n  +  a 
proper  fraction. 

For  example :  The  number  4570  is  between  10^  and  10* ; 
hence  its  logarithm  is  3  -f  a  proper  fraction.  The  num- 
ber .00457  is  between  10~^  and  10~^ ;  hence  its  logarithm 
is  —  3  +  a  proper  fraction. 

It  follows,  from  these  two  principles,  that  in  the  con- 
struction of  a  table  of  logarithms  to  the  base  10  : 


Numbers. 

Logarithms. 

.0001 

-4 

.001 

-3 

.01 

-2 

.1 

-1 

1 

0 

10 

1 

100 

2 

1000 

3 

10000 

4 

TO   OBTAIN  A  LOGARITHM. 


41 


(1)  The  mantissae  need  be  given  but  once  for  numbers 
having  the  same  figures  in  the  same  order. 

(2)  The  characteristics  need  not  be  given,  but  can  be  sup- 
plied by  the  following  rule  : 

Make  the  characteristic  equal  to  the  number  of  places  by  which 
the  first  significant  figure  of  the  number  is  removed  from  the 
units  place :  positive  if  the  first  significant  figure  is  to  the  left  of 
the  decimal  point,  negative  if  to  the  right. 


THE  USE  OF  THE  TABLE  OF  LOGARITHMS. 

22.  To  find  the  logarithm  of  a  number.  —  The  mantissae  of 
the  logarithms  of  numbers  from  0  to  1000  are  printed  on 
pages  viii  and  ix  of  the  tables  at  the  back  of  the  book. 
If  any  mantissa  is  printed  without  the  decimal  point,  a 
figure  of  the  mantissa  has  also  been  omitted ;  the  decimal 
point  and  missing  figure  will  be  found  above  and  to  the  left 
of  the  figures  in  the  table. 

The  first  two  significant  figures  of  the  number  will  be 
found  in  the  column  marked  N,  and  the  third  significant 
figure  at  the  top  of  the  columns.  If  a  given  number  has 
but  one  significant  figure,  two  zeros  are  added ;  if  but  two 
significant  figures,  one  zero  is  added.     Thus : 


PAGE  VIII. 

PAGE 

IX. 

Line. 

Number. 

Logarithm. 

Line. 

Number. 

Logarithm. 

11 

2.00 

0.3010 

11 

.065 

.8129  -  2 

14 

23.0 

1.3617 

14 

.068 

.8325  -  2 

17 

2.61 

0.4166 

17 

.717 

.8555  —  1 

20 

29.4 

1.4683 

20 

.074 

.8692  -  2 

23 

32.7 

1.5145 

23 

.777 

.8904  -  1 

36 

456 

2.6590 

36 

.009 

.9542  -  3 

38 

4.70 

0.6721 

38 

.920 

.9638  -  1 

42         EXPLANATION  OF  THE  TABLES. 

To  find  the  logarithm  of  a  number  of  four  significant 
figures,  the  assumption  is  made  that  small  changes  in  the 
numbers  produce  proportional  changes  in  the  logarithms.  This 
assumption  is  only  approximately  true;  that  it  is  nearly 
true  may  be  seen  from  an  inspection  of  the  table.  The 
error  in  the  assumption  will  not,  in  general,  affect  the  fourth 
figure  of  the  mantissa  of  the  logarithm. 

Thus,  let  m  stand  for  mantissa  and  log  for  logarithm. 
Then 

m  log  200.2  =  m  log  200  +  .2  (m  log  201  -  m  log  200)  ; 

=  .3010  +  .2  (.3032  -  .3010). 

Therefore  log  of  200.2  =  2.3014. 
m  log  .02307  =  m  log  230  +  .7  (m  log  231  -  m  log  230) ; 
=  .3617  +  .7  (.3636  -  .3617). 
Therefore  log  of  .02307  =  .3630  -  2. 

Given  a  logarithm,  to  find  the  corresponding  number.  —  Let 
log  iV=  1.4997.  Then  (page  viii,  line  22)  the  significant 
figures  of  N  are  316.  The  characteristic  indicates  that  the 
figure  3  is  one  figure  to  the  left  of  the  units  place ;  hence 
the  figure  1  is  in  the  units  place  and  N=  31.6. 

Let  log  N  =  .5453  —  2.  The  significant  figures  of  J^  are 
351,  and  since  the  3  is  two  figures  to  the  right  of  the  units 
place,  N=  0.0351. 

Let  log  N=  2.3014.     Then 

jsr=  200  +  -^Qi^  -  -^QiQ  ^  200.2. 

.3032  -  .3010 


Let  log  N  =  .3630  -  2.     Then 

3630  - .; 
.3636  -  .3617 


Q23Q -3630 -.3617  ^  Q2307. 


INTERPOLATION.  43 

By  assuming  either  column  in  this  table  find  the  other. 


Number. 

Logarithm. 

Number. 

Logarithm. 

10.62 
1.384 
1886 
273.1 

1.0261 
0.1411 
3.2755 
2.4363 

4.426 
56.81 
6.282 
733.5 

0.6460 
1.7544 
0.7981 
2.8654 

Other  methods  of  writing  negative  characteristics.  — To  avoid 
printing  negative  characteristics,  it  is  usual  to  arbitrarily 
increase  the  characteristic  by  10.  Thus,  log  of  .01657  = 
.2193  —  2  will  sometiriies  be  written  8.2193,  which  is  the 
logarithm  of  165,700,000,  or  just  10,000,000,000  times  as 
great  as  the  real  value.  In  such  a  case,  L  will  be  used 
instead  of  log.     Thus 

log  .01657  =  8.2193 -10  and  X.  01657  =  8.2193. 

Suppose  the  cube  root  of  .01657  is  desired. 

Since  log  .01657  =  .2193  -  2  or  8.2193  -  10, 


log  V  .01657  =  .0731  -    .6667  or  2.7398  -  3.3333. 

To  avoid  this  negative  mantissa,  the  logarithm  of  .01657 
is  arranged  before  division,  thus : 

log  .01657  =  1.2193  -  3  or  28.2193  -  30. 


Therefore  log  v  .01657  =  .4064  - 1  or  9.4064  -  10. 

That  is :  In  taking  the  root  of  a  proper  fraction  by  loga- 
rithms, see  that  the  negative  part  of  the  logarithm  is  divisi- 
ble by  the  index  of  the  root. 

Negative  numbers.  —  The  suffix  n  placed  after  a  logarithm 
indicates  that  the  number  corresponding  to  the  logarithm  is 
negative.  If,  in  obtaining  the  product  of  several  numbers, 
there  is  an  odd  number  of  negative  factors,  the  suffix  n  will 


44 


EXPLANATION  OF  THE   TABLES. 


occur  an  odd  number  of  times ;  therefore,  by  the  principles 
of  algebra,  the  logarithm  of  the  product  should  have  the  suf- 
fix n.  But  if  there  is  an  even  number  of  negative  factors, 
the  suffix  n  will  appear  an  even  number  of  times ;  therefore 
the  logarithm  of  the  product  will  not  contain  the  suffix  n. 

In  a  similar  manner,  the  rules  of  algebra  will  serve  as  a 
guide  in  the  use  of  the  suffix  n,  in  the  operations  of  division, 
involution,  and  evolution. 


1.  Compute 

X  =  46.12  X  .7827. 

log  46.12=    1.6639 
log  .7827=    9.8936-10 
log  X  =  11.5575  —  10 
a;=:36.1 

2.  Compute 

X  =  525.1  --  78530. 

log  525.1  =  12.7202  -  10 
log  78530=    4.8950 

loga;=    7.8252  -  10 
X  =  0.006687 

3.  Compute 

X  =  5850  H-  0.04754. 

log  5850  =  13.7672  -  10 
log  .04754  =    8.6771  -  10 
log  X  =    5.0901 
X  =  123100 


4.  Compute 


4 


fl62.4  X  238.9  x  115.5 


516.8 
log  162.4  =  2.2106 
log  238.9  =  2.3782 
log  115.5  =  2.0626 

6.6514 
log  516.8  =  2.7133 

log  X'  =  3.9381 
log  X  =  1.9690 
X  =  93.12 

5.  Compute 

3/129.97  X  .0499 


=4 


278.7 
log  129.97=    2.1138 
log  .0499=    8.6981-10 

10.8119  -  10 
log  278.7=    2.4451 

log  0^  =  28.3668  -  30 
logx=    9.4556-10 
X  =  0.2855 


Exercises.  —  Use  problems  on  page  10. 


EXERCISES. 


45 


III.     LOGARITHMS  OF  FUNCTIONS. 

23.  The  table,  pages  x  to  xvi,  Logarithms  of  Functions, 
contains  the  logarithms  of  the  natural  functions  of  angles 
from  0°  to  90°.  To  avoid  printing  negative  characteristics, 
the  characteristics  of  logarithms  of  numbers  less  than  1 
have  been  printed  10  too  large  (indicated  by  the  letter  L 
in  the  first,  second,  and  sixth  columns).  Hence  —10  will 
be  understood  after  such  logarithms. 

Just  as  in  the  other  tables,  any  number  printed  without 
the  decimal  point  will  have  also  one  or  more  figures  missing, 
which  figures  will  be  found  above  and  to  the  left. 

The  method  of  interpolation  is  the  same  as  in  the  use  of 
the  table  of  natural  functions. 

Exercises. — Copy  a  value  in  any  row  in  the  following 
table;  then  find  each  of  the  remaining  values  in  the  row: 


sin    18°   5' 

sin  161°  55' 

9.4919 

cos   71°  55' 

cos  288°   5' 

sin    26°   6' 

sin  153°  54' 

9.6434 

cos   63°  54' 

cos  296°   6' 

sin    25°  52' 

sin  154°   8' 

9.6398 

cos   64°   8' 

cos  295°  52' 

sin  196°  14' 

sin  343°  46' 

9.4465  n 

cos  106°  14' 

cos  253°  46' 

sin  251°  31' 

sin  288°  29' 

9.977071 

cos  161°  31' 

cos  198°  29' 

tan  112°  14'  tan  292°  14 

tan  101°  46'  tan  281°  46 

tan  103°  17'  tan  283°  17 

tan  177°  37'  tan  357°  37 

tan   34°  44'  tan  214°  44 

sec   22°  13'  sec  337°  47 

sec  142°  23'  sec  217°  37 

sec   76°   8'  sec  283°  52 

sec    96°  26'  sec  263°  34 

sec   69°  36'  sec  290°  24 


0.3885  n  cot  157°  46'  cot  337°  46' 

0.6813  ?i  cot  168°  14'  cot  348°  14' 

0.6269  n  cot  166°  43'  cot  346°  43' 

8.6193  n  cot   92°  23'  cot  272°  23' 

9.8409  cot   55°  16'  cot  235°  16' 

0.0335  esc   67°  47'  esc  112°  13' 

0.2144  n  CSC  232°  23'  esc  307°  37' 

0.6204  CSC   13°  52'  esc  166°   8' 

0.9506  n  CSC  186°  26'  esc  353°  34' 

0.4577  CSC   20°  24'  esc  159°  36' 


CHAPTER   III. 

RIGHT    TRIANGLES. 

24.  The  primary  object  of  trigonometry  is  the  investiga- 
tion of  the  methods  of  computing  the  numerical  values  of 
the  sides  and  angles  of  plane  triangles.  When,  from  any 
data,  such  numerical  values  have  been  obtained,  the  triangle 
will  be  said  to  be  solved. 

With  reference  to  the  number  of  possible  answers,  a 
problem  is  said  to  be  indeterminate,  ambiguous,  determinate,  or 
impossible: 

(1)  When  there  is  an  infinite  number  of  answers,  the 
problem  is  indeterminate.  An  algebraic  illustration  is : 
find  X  and  y  from  the  equation  x  —  y  =  ^. 

(2)  When  there  is  more  than  one  answer,  but  not  an 
infinite  number  of  answers,  the  problem  is  ambiguous.  For 
example :  find  x  from  the  equation  a^  +  6  =  5  ic. 

(3)  When  there  is  one  answer  and  but  one  answer  for 
each  unknown,  the  problem  is  determinate.  For  example : 
find  X  and  y  from  the  equations  x-{-  y  =  7  and  x  —  y  =  3. 

(4)  When  there  is  no  answer,  the  problem  is  impossible. 
For  example :  find  x  from  the  equation 

2  x—5  =  negative  square  root  of  (ot^  —  7). 

In  trigonometry,  negative  and  imaginary  answers  are,  in 
general,  rejected,  as  not  indicating  true  solutions. 

It  is  known  from  geometry  that  at  least  three  indepen- 
dent conditions  are  necessary  to  make  the  solution  of  a 

46 


RELATIONS  OF  ANGLES  AND  SIDES.  47 

triangle  determinate.*     In  most  cases,  three  such  conditions 
are  sufficient. 

Two  classes  of  triangles  will  be  considered,  —  right  and 
oblique.  Any  oblique  tri- 
angle can  be  divided  into 
two  right  triangles  by  let,- 
ting  fall  from  any  vertex 
a  perpendicular  upon  the 
opposite  side.  Hence  the 
solution  of  all  triangles 
depends  upon  the  solution 
of  right  triangles.  Fig.  27. 

25.  In  a  right  triangle,  let  A  denote  the  value  of  angle  A^ 
B  denote  the  right  angle,  and  C  the  value  of  angle  C.  Also 
let  a  and  c  denote  the  lengths  of  the  sides  of  the  triangle 
opposite  A  and  C,  respectively,  and  b  denote  the  length  of 
the  hypotenuse. 

Relations  of  angles.  —  In  a  right  triangle,  one  of  the  three 
conditions  necessary  to  determine  its  size  and  its  shape  is 
explicitly  given.     For  the  angle  B  =  90°.     Therefore 

A+C=90°;  A  =  90°-C',  C  =  90°-A. 

Hence,  when  either  acute  angle  of  a  right  triangle  is  given, 
the  other  acute  angle  is  readily  found. 

Relations  of  sides.  —  The  relation  existing  between  the 
sides  of  a  right  angle  is  well  known.     It  is 


6=  Vc^  +  a';  c=  V(?>  +  a)(6-a);  a=  V(6  +  c)  (6  -  c). 

Hence,  if  any  two  of  the  sides  of  a  right  triangle  are 
given,  the  other  side  can  be  computed. 

*  This  is  expressed  by  saying  that  the  triangle  is  of  triple  mani- 
foldness. 


48  RIGHT  TRIANGLES. 

Compute  a,  given  6=  145,  c  =  143. 


a  =  V(145  -f  143)  (145  -  143)  =  24. 
Compute  c,  given  a  =  399,  b  =  401. 


c  =  V(401  +  399)  (401  -  399)  =  40. 
Exercises.  —  Given  b  and  c ;  find  a  in  each  of  the  following ; 


b 

c 

a 

919.2 

918.94 

21.90 

865.1 

860.74 

86.90 

888.9 

888.84 

10.34 

150.0 

149.50 

12.29 

525.1 

523.84 

36.35 

Relations  of  sides  and  angles.  —  Assume  AB  to  be  an  initial 
line,  c  the  abscissa,  a  the  ordinate,  and  b  the  distance  of  C. 


tan  A 


sin  A 


The  rules  on  page  19  give 

Any  required  side  =  product  of  adjacent  side  and  adjacent 
function. 


SOLUTIONS   USING  NUMBERS.  49 

Any  function  oi  A  =  quotient  of  the  adjacent  side  by  the  next 
side. 

If  it  is  desired  to  use  the  angle  C  in  a  computation,  the 
letters  a  and  c,  A  and  (7,  can  be  interchanged  in  Fig.  28, 
and  the  rules  applied  to  the  resulting  figure. 


SOLUTIONS. 

26.  1.   Given  6  =  120,  A  =  11°  25';  find  a. 

The  horizontal  cross-line  in  Fig.  28  connects  the  given  side 
b  and  the  required  side  a ;  hence  by  the  rule,  a  =  6  sin  A. 

That  is,  a  =  120  sin  11°  25' 

=  120  X  .1979  =  23.75. 

2.  Given  c  =  50.7,  ^  =  57°  7';  find  6. 

Here  we  use  the  vertical  cross-line,  since  that  connects 
the  given  side  and  the  required  side.     Hence  b  =  c  sec  A. 

That  is,  6  =  50.7  sec  57°  7' 

=  50.7  X  1.8419  =  93.38. 

3.  Given  a  =  71,  A  =  65°  32' ;  find  c. 
The  oblique  cross-line  gives 

c  =  a  cot  ^  =  71  cot  65°  32' 
=  71  X  .4550  =  32.3. 

4.  Given  6  =  75,  a  =  43.8 ;  find  A. 

Here  A  can  be  found  by  means  of  its  sine  or  its  cosecant. 
As  the  division  by  75  is  slightly  easier,  we  select  the  sine. 

sin ^  =  —  =.5840;    ^  =  35° 44'. 
75 


60 


RIGHT  TRIANGLES. 


5.   Given  a  =  849,  c  =  920 ;  find  A. 

Here  the  oblique  cross-line  connects  the  two  given  sides. 
As  920  is  the  easier  divisor, 

849 
920 


tan^ 


.9228;  ^  =  42°  42'. 


6.   Given  b  =  401,  c  =  399 ;  find  A. 

Here  cos  .4  =  |^  =  .9950. 

401 

In  this  solution,  A  cannot  be  determined  with  much  pre- 
cision as  the  table  gives  5°  43',  5°  44',  or  5°  45'. 
Hence  find  a  first ;  a  =  40.     Then 


csc^  =  ^  =  10.025;  A 


5°  43^'. 


It  will  be  noticed  that  the  cosines  and  secants  of  small 
angles  change  slowly ;  and  that  sines  and  cosecants  of  angles 
near  90°  change  slowly.  When  precision  is  desired,  such 
functions  of  these  angles  are  avoided. 

The  following  table  gives  the  three  sides  and  one  angle 
of  eight  different  right  triangles.  Assume  any  two  parts 
of  any  one  of  these  triangles  and  compute  another  part. 


a 

b 

c 

A 

4 

5 

3 

53°  8' 

12 

13 

5 

67°  23' 

24 

25 

7 

73°  44' 

20 

29 

21 

43°  36' 

12 

37 

35 

18°  55' 

28 

53 

45 

31°  53' 

60 

109 

91 

33°  24' 

84 

205 

187 

24°  11' 

APPLICATIONS. 


61 


Fig.  29. 


Fig.  30. 


PROBLEMS. 

[Answers  to  these  problems  will  be  found  on  page  177.] 

27.  1.  A  tower,  BC,  stands  on  flat  horizontal  ground ;  a 
distance  of  250  feet  (c)  was  measured  from  the  base  of  the 
tower  to  a  point  A.  At  this  point  the  angle 
between  the  horizontal  {AE)  and  the  visual 
ray  (^0)  to  the  top  of  the  tower  was  found 
to  be  ^  =  22°  20'.*  From  these  data,  com- 
pute the  height  of  the  tower. 

2.  A  lighthouse,  BC,  is  87  (a) 
feet  high.  From  the  top  C,  the 
angle  between  a  horizontal  line 
CD  and  the  visual  ray  to  a  boat 
A  is  found  to  be  12°  40'.t  From 
these  data  compute  the  distance 
of  the  boat  from  the  base  of  the 
lighthouse. 

3.  Two  forces,  AB  of  240  lbs.  and  AD  of  100  lbs.,  act 
at  right  angles  upon  the  same  point  of  a  body.  Find  the 
angle  which  their  resultant  AC  makes  with  the  larger 
force. 

4.  The  length  of  the  projection  of  ^C  on  a  line  AB  is 
23  inches ;  the  angle  which  AC  makes  with  this  projection 
is  38°.     Find  the  length  of  AC. 

5.  A  cannon  ball,  fired  in  the  air  at  an  angle  of  33°  3' 
with  the  horizon,  passes  over  880  feet  in  one  second.  How 
high  is  it  at  the  end  of  the  second  ?  t 

*  This  angle  is  called  the  angle  of  elevation  of  the  top  of  the  tower, 
as  seen  from  A. 

t  This  angle  is  called  the  angle  of  depression  of  the  boat  as  seen 
from  C. 

t  This  increase  of  height  per  second  is  called  the  vertical  component 
of  the  velocity. 


52  RIGHT  TRIANGLES. 

« 

6.  In  order  to  measure  the  breadth  of  a  river,  a  base  line 
AB  was  measured  along  the  bank,  the  point  B  being  directly 
opi)osite  a  tree  C  on  the  other  bank.  The  base  line  AB  was 
96  (c)  feet,  and  the  angle  BAG  was  found  to  be  52°  30'  (A). 
What  is  the  breadth  of  the  river  ?  * 

7.  When  the  rays  of  the  sun  make  with  the  horizontal 
plane  an  angle  of  53°  35'  (^),t  a  vertical  pole  throws  a 
shadow  11.8  (c)  feet  long.  Find  the  length  of  the  pole  if 
its  lower  end  is  1  foot  (d)  underground. 

8.  From  a  window  of  a 
house,  the  angle  of  ele- 
vation of  the  top  of  a 
church  tower  on  the  op- 
posite side  of  the  street 
was  56°36'(^),  and  the 

angle    of    depression     of  a^^- 

the  base  of  the  tower 
was  10°  24'  (A)-  The 
street  being  60  feet  (c) 
wide,  find  the  height  of 
the  steeple. 

9.  A  house  is  24  feet  (a)  high.  From  the  top  of  the 
house,  the  angle  of  elevation  of  a  church  tower  on  the 
opposite  side  of  the  street  is  36°  54'  (A),  and  the  angle 
of  depression  of  the  base  of  the  tower  is  18°  56'  (Ai).  Find 
the  height  of  the  steeple. 

10.  From  the  top  of  a  hill,  the  angles  of  depression  of  the 
top  and  of  the  bottom  of  a  house  44  feet  (a)  high  are  found 
to  be  28°  4'  {A)  and  43°  36'  (^i),  respectively.  Find  the 
height  of  the  hill  above  the  ground  on  which  the  house  stands. 

*  A  segment  drawn  across  an  angle  is  said  to  subtend  the  angle. 
Thus  in  a  triangle  ABC,  BC  subtends  the  angle  A. 

t  The  angle  which  the  sun's  rays  make  with  the  horizon  is  equal  in 
degrees  to  the  arc  which  measures  the  altitude  of  the  sun. 


Jc. 


Fig.  31. 


c 


/i 


APPLICATIONS.  53 

11.  From  the  top  of  a  bluff  overhanging  a  river,  the 
angle  of  elevation  of  the  top  of  a  tree  on  the  other  side  of 
the  river  is  5°  22'  (^),  and  the  angle  of  depression  of  the 
bottom  of  the  tree  is  3°  40' 
(yli).  The  tree  is  known 
to  be  79  feet  (a)  high. 
What  is  the  breadth  of  the 
river  ? 

12.  A  man   sailing  due                           y           / 
north  observes  two  objects                     y'             / 
directly  west.     After  sail-                ,'^                 /' 
ing  12  miles  (a),  the  direc-          ^^^                    / 
tions  of  the  objects  make      X_ /_ 


B 


angles  of  22°  38'  (C)*  and  ^ 

of  16°  16'   ((7i)   with  the  ^'''•^^• 

ship's  course.     How  far  apart  are  the  objects? 

13.  At  a  horizontal  distance  of  71  feet  (c)  from  the  middle 
of  the  base  of  a  church  tower,  the  angle  of  elevation  of  the 
top  of  the  tower  =  24°  27'  {A),  and  of  the  top  of  the  steeple 
=  66°  10'  (A)-     Find  the  height  of  the  steeple. 

14.  Two  observers,  570  (c)  feet  apart  and  facing  each 
other,  observe  a  balloon  in  the  vertical  plane  passing 
through  them,  at  angles  of  elevation  of  67°  22'  {A)  and 
61°  56'  (^i),  respectively.  What  is  the  height  of  the  bal- 
loon above  the  ground  ? 

15.  The  angle  of  elevation  of  a  house  on  the  bank  of  a 
creek,  observed  from  the  opposite  bank  =  48°  35'  {A).  The 
observer  walks  straight  back  from  the  river  a  distance  of 
43.15  feet  (c),  and  the  angle  of  elevation  is  then  33°  24'  (^i). 
Find  the  breadth  of  the  creek. 

*  The  angle  which  a  horizontal  line  makes  with  the  meridian  is  called 
the  bearing  of  the  line.  Thus  \i  BC  is  a  meridian,  and  a  point  A  is 
22°  38'  west  of  south  from  a  point  C,  the  bearing  of  A  from  C  is 
written  S.  22°38'  W. 


54  RIGHT  TRIANGLES. 

16.  The  angles  of  elevation  of  the  top  of  a  steeple  at 
distances  of  50  feet  (c)  and  450  feet  (cj)  are  found  to  be 
complementary.     Find  the  height  of  the  steeple. 

17.  A  man  at  the  top  of  a  lighthouse  observes  a  boat 
heading  straight  for  the  lighthouse ;  the  angle  of  depression 
is  34°  40'  (A) ;  ten  minutes  (t)  later,  it  is  55°  35'  (A).  How 
soon  will  the  boat  reach  the  lighthouse  ? 

18.  Kain  falls  at  an  angle  of  75°  30'  (A)  with  the  ground. 
The  wind  blows  at  right  angles  to  a  wall,  17.5  feet  (a)  high. 
How  far  from  the  wall  can  a  man  stand  without  getting 
wet,  his  height  being  5.5  feet  (h)  ? 

19.  From  the  end  of  the  diameter  of  a  circle,  a  chord  is 
drawn,  the  length  of  which  is  n  times  the  diameter.  Find 
the  angle  between  the  chord  and  the  diameter.     Take  n=.62. 

20.  Find  the  altitude  of  the  sun  if  the  height  of  a  man  is 
n  times  the  length  of  his  shadow.     Take  n  =  .74. 

21.  From  the  top  of  a  lighthouse  221  feet  (a)  above  the 
ocean,  two  boats  are  ob- 

served,  one  due  west  hav-  ,^ 

ing  an  angle  of  depression 

of  67°  37'   (A),  the  other 

due  south  having  an  angle 

of    depression    of    74°  49' 

(Z>).      Find   the   distance 

between  the  boats.  A^'^--- 

22.  To      measure      the 
height    BC    of    Si    tower 
across   a  river,  the  tower  ^^^         /'  ^^^ 
being  directly  opposite  a  ^^J"^^ 
point  A,  a   distance   AD  ^ 

=  500  feet  (d)  was  meas-  ^'^-  ^^• 

ured  along  the  bank.  At  A  the  angle  of  elevation  of  the 
top  of  the  tower  was  13°  58'  (^),  and  at  D  the  elevation  was 
10°  (i>).     Find  the  height  of  the  tower. 


AREA. 


65 


THE  AREA  OF  A  RIGHT  TRIANGLE. 


'A. 


28.  From  the  right  angle  of  the  triangle,  drop  a  perpen- 
dicular upon  the  hypotenuse.     Denote  the  length  of  this 

c 


perpendicular  by  jp,  and  the  segments  of  the  hypotenuse  by 
d  and  e.     Denote  the  area  by  K. 
By  geometry,  K^\ac  =  \  bp. 

Exercises.  —  Prove  the  following  relations  involving  the 
area  of  a  right  triangle: 


(7)  4ir2  =  a2(62_a3). 

(8)  4.IP=c'(b'-(^, 

(9)  4.K  =h^-(a-  cf. 

(10)  16K^-4:a:'(PK^=a''(P. 

(11)  2K=d\tsinA-\-t3i.n^A). 

(12)  8^=(a  +  c)2-(a-c)l 

(13)  4  Jr=(a  4-  6  +  c)  (a  -  6  +  c). 

(14)  8  K=  [(a  +  cy  +  (a  -  c)^]  sin  2  A 

If  K  is  eliminated  between  any  two  of  these  relations,  a 
relation  between  the  parts  of  the  triangle  is  obtained.  Thus 
if  the  2d  and  9th  relations  are  used,  there  is  obtained 

b'sm2A  =  b^-(a-cy. 


(1)  2K  =a2cotA 

(2)  4:K  =b'sm2A. 

(3)  2K  =cHanA 

(4)  K  :=:p^csc2A. 

(5)  4.K  =(c2-a2)  tan  2  A 

(6)  4:K^=bH-b^d\ 


b6 


RIGHT  TRIANGLES. 


23.  A  man  having  traveled  32  miles  in  a  straight  line, 
finds  that  he  has  made  17.3  miles  more  toward  the  north 
than  toward  the  east.    What  direction  has  he  been  traveling  ? 

24.  A  triangular  field  is  at  the  corner  of  two  roads  which 
form  a  right  angle.  The  area  of  the  field  is  32,130  sq.  ft., 
and  the  total  length  of  the  fence  enclosing  it  is  918  feet. 
Find  the  frontage  on  each  road. 

25.  A  triangular  field  is  at  the  corner  of  two  roads  which 
form  a  right  angle.  The  area  of  the  field  is  six-elevenths 
of  an  acre,  and  the  length  of  the  fence  along  the  two  roads 
is  487  feet  (the  total  front  fence).  Find  the  length  of  the 
fence  on  the  back  of  the  field. 

SOLUTIONS  USING  LOGARITHMS. 


29.  1.   Given 

3.   Given  a  =  .129,    c  =  .089. 

a  =  152,' ^  =  18°  25'. 

Find  A  and  b. 

Find  b  and  c. 

La           9.1106 

log  CSC  ^0.5004 

Lc            8.9494 

log  a         2.1818 
log  cot  ^0.4776 

^=55°  24'    log  tan  ^0.1612 
log  sec  ^  0.2458 

5=481.1      log  6         2.6822 

Lc            8.9494 

c=456.5      logc         2.6594 

&  =  .1567      Lb           9.1952 

2.   Given 
c  =  .3917,   ^  =  65°  14'. 
Find  a  and  b. 

log  tan  ^0.3360 
Lc           9.5930 
log  sec  ^0.3779 

4.   Given 

&  =  86.53,  ^  =  56°  3'. 

Find  a  and  c. 

LsinA    9.9188 
log  b         1.9372 
LgosA    9.7470 

a=.8492      La           9.9290 

a=71.78    log  a         1.8560 

5=.9352      Lb           9.9709 

c=48.33    logc         1.6842 

SOLUTIONS  USING  LOGARITHMS.  57 

5.   Given  a  =  50.7,  h  =  93.4.     Find  the   unknown  parts 
and  the  area.     Check  the  work. 


h  =  93.4 

log  6 

1.9703 

a  =  50.7 

log  a 

1.7050 

^  =  32°  53' 

log  CSC  A 

0.2653 

.-.  C=57°7' 

log  cot  A 

0.1894 

log  a 

1.7050^ 

c  =  78.42 

logc 

1.8944 

X.5 

9.6990 

K=  1988 

log^ 

3.2984 

6  +  a  =  144.1 

log  (b  +  a) 

2.1587 

6  -  a  =  42.7 

log  (b  —  a) 

1.6304 

Check 

:  logc^ 

3.7891 

6.   Given  b  =  31.6,  c  =  31.4 ;  find  a  and  A. 

Here  the  hypotenuse  and  side  are   given  nearly  equal. 


By  (11), 

tan  ^A=  CSC  A 

a         ^6  +  c 

L(b-c)    9.3010 

log  (6 +  c)  1.7993 

loga^         1.1003 

itan^  1^7.5017 

a  =  3.549 

log  a          0.5501 

i^  = 

3n3f;   A  = 

6° 

27'        itan^^  8.7508 

Exercises.  —  Assume  any  two  parts  and  find  the  remain- 
ing parts  in  the  following  right  triangles: 

1.  a  =  250  6  =  627  c  =  575  ^  =  23°  30'  A' =  71880 

2.  a  =  13.13  6  =  21.94  c  =  17.58  ^  =  36°  45'  /f=  115.4 

3.  a  =  41.54  6  =  42  c  =  6.208  ^1  =  81°  30'  7^=1289 

4.  a  =  .3864  6  =  4.65  c  =  4.634  A=    4°  46'  7f  =  .8953 


58  RIGHT  TRIANGLES. 

PROBLEMS. 
30.   Solve  the  following  problems,  using  logarithms : 

26.  A  wall  is  surrounded  by  a  ditch.  A  ladder  placed 
with  its  foot  at  the  edge  of  the  ditch  just  reaches  the  top 
of  the  wall  on  the  other  side  of  the  ditch.  The  length  of 
the  ladder  was  33.7  feet,  and  the  angle  which  it  formed 
with  the  horizontal  plane  was  24°.  Find  the  height  of  the 
wall  and  the  width  of  the  ditch. 

27.  Two  forces  at  right  angles,  one  of  which  is  7.25 
pounds,  are  acting  at  the  same  point  of  a  body.  Their 
resultant  divides  the  angle  between  them  into  parts  pro- 
portional to  3  and  5,  the  smaller  angle  being  next  to  the 
given  force.  Calculate  the  other  component  force  and  the 
resultant  of  the  two  forces. 

28.  The  diagonal  of  a  sheet  of  paper  is  16.5  inches,  and 
it  makes,  with  the  shorter  side,  an  angle  of  57°  57'.  Calcu- 
late the  sides  of  the  sheet. 

29.  On  flat  horizontal  ground  154  feet  from  the  foot  of 
a  tower,  the  angle  of  elevation  of  the  top  was  27°  10'.  The 
axis  of  the  telescope  was  five  feet  from  the  ground.  Ee- 
quired  the  height  of  the  tower. 

30.  The  latitude  of  Philadelphia  is  40°  IN".  Taking  the 
radius  of  the  earth  at  Philadelphia  to  be  3960  miles,  find 
the  radius  and  the  length  of  1°  of  the  parallel  passing 
through  Philadelphia. 

31.  The  circumference  of  the  meridian  passing  through 
Paris  is  40,000  kilometers.  The  length  of  1°  of  the  parallel 
passing  through  Paris  =  73.13  km.  Find  the  latitude  of 
Paris,  considering  the  earth  a  sphere. 

32.  Two  forces,  P  =  7.25  lbs.,  and  Q  =  10.3  lbs.,  act  at 
right  angles  on  a  body.  Calculate  the  resultant  and  the 
angle  that  it  makes  with  P. 


APPLICATIONS.  69 

33.  Calculate  the  velocity  per  second  of  a  point  in  40° 
latitude  as  the  earth  rotates  on  its  axis.  Take  the  radius 
of  the  earth  =  3960  miles. 

34.  A  frigate  is  4  miles  west  from  a  steamer  which  is 
sailing  north  at  the  rate  of  22  miles  an  hour.  How  far  in 
front  of  the  steamer  must  the  gunner  on  the  frigate  aim 
in  order  to  strike  the  steamer,  if  the  average  velocity  of  the 
cannon  ball  for  that  distance  is  800  feet  per  second  ? 

35.  A  square  is  inscribed  in  another  square ;  the  parts  of 
the  sides  of  the  outer  square  made  by  the  corners  of  the 
inner  square  are  965  (m)  and  603  (?i).  Find  the  angle  made 
by  the  diagonal  of  the  inner  square  and  the  side  of  the 
outer  square. 

36.  An  observer  sees  a  cloud  in  the  southwest  at  an 
angle  of  elevation  of  43°  35'.  Another  observer  2526  feet  (b) 
south  of  the  first  sees  the  same  cloud  in  the  northwest. 
How  high  is  the  cloud  ? 

37.  A  pedestal  12  feet  high  (a)  supports  a  column  13  feet 
high  (h).  An  observer,  whose  eye  is  in  the  same  horizontal 
plane  with  the  base  of  the  pedestal,  sees  the  pedestal  and 
column  each  subtending  the  same  angle.  What  is  the  hori- 
zontal distance  of  the  observer  from  the  pedestal  ? 


CHAPTER   IV. 

THE  ISOSCELES  TRIANGLE  AND  THE  REGULAR  POLYGON. 

31.   The  isosceles  triangle  can   be 

divided  into  two  equal  right  triangles 
by  drawing  an  altitude  upon  the  un- 
equal side.  Hence  the  formulae  for 
right  triangles  will  apply  to  the  half- 
triangle  ABiC.     That  is, 

p  =  bsmA  =  ^c  tan  A. 
b  =  ^cseGA=pcsGA. 
c  =  2p  cot  A  =  2h  cos  A. 
Area  =  K=  ^Gp  =  ^hG  sin  A 

=  \h''^m2A  =  \cH2iXiA. 

In  these  relations,  A  may  be  changed  to  ^  (7  and  the  co- 
functions  taken. 

Exercises.  —  Assume  any  two  parts  in  any  row  of  the  fol- 
lowing table  and  compute  the  remaining  parts  in  that  row : 


p 

h 

c 

A 

K 

16.80 

17.00 

5.200 

8ri2' 

43.68 

1.960 

2.275 

2.310 

59°  29' 

2.264 

70.00 

101.5 

147.0 

43°  36' 

5145 

10.80 

33.30 

63.00 

18°  ob' 

340.2 

3.179 

4.264 

5.684 

48°  12' 

9.034 

25.00 

62.70 

115.0 

23°  30' 

1438 

.5878 

1.000 

.1618 

36°  00' 

.4755 

18.08 

83.50 

163.0 

12°  30' 

1474 

60 


APPLICATIONS.  61 


PROBLEMS. 


32."  38.  Two  trains  start  at  the  same  station  at  the  same 
time  on  tracks  making  an  angle  of  67°  30'.  Each  goes  with 
a  velocity  of  27  miles  per  hour.  In  how  many  minutes 
will  the  distance  between  them  be  equal  to  5  miles? 

39.  The  chord  of  a  circle  is  27  inches,  and  the  angle  at 
the  center  subtended  by  the  chord  is  116°  12'.  Find  the 
radius. 

40.  The  angle  at  the  vertex  of  an  isosceles  triangle  is 
87°  13',  and  the  sum  of  the  two  unequal  altitudes  is  60. 
Find  the  base. 

41.  The  ratio  of  the  altitude  from  the  vertex  of  an  isos- 
celes triangle  to  one  of  the  equal  altitudes  is  1.2,  and  each 
of  the  equal  sides  =  5.3.  Find  the  base  angle  and  the 
area. 

42.  From  a  point  without  a  circle  12  feet  (d)  from  the 
center,  two  tangents  are  drawn,  making  with  each  other  an 
angle  of  25°  3'  (A).  Calculate  the  area  of  an  equilateral 
triangle  inscribed  in  this  circle. 

43.  From  a  point  without  a  circle  23.2  feet  from  the  cen- 
ter, two  tangents  are  drawn.  The  radius  of  the  circle  equals 
9.6  feet.  Calculate  the  side  of  an  equilateral  triangle  whose 
area  is  equal  to  that  of  the  figure  bounded  by  the  two  tan- 
gents and  by  the  smaller  arc  included  between  the  points  of 
contact. 

44.  The  slant  height  of  a  right  cone  is  13  feet,  and  it 
makes  with  the  plane  of  the  base  an  angle  of  67°  23'.  Find 
the  convex  surface  of  the  cone. 

45.  When  a  cone  was  cut  through  its  axis,  the  area  of  the 
section  was  found  to  be  10  sq.  inches  and  the  angle  at  the 
vertex  40°  30'.     Find  the  convex  surface  of  the  cone. 


62  ISOSCELES  TRIANGLE  AND  REGULAR  POLYGON. 

46.  The  convex  surface  of  a  cone  =  80  sq.  inches  (S),  and 
the  angle  at  the  vertex  of  the  principal  section  =  39°  13'  (C). 
Find  the  volume  of  the  cone. 

47.  When  the  convex  surface  of  a  right  cone  was  spread 
out  in  a  plane,  it  was  found  to  be  a  sector  of  a  circle  whose 
central  angle  was  108°  36',  and  the  chord  subtending  the 
corresponding  arc  was  found  to  be  12  inches.  Calculate 
the  volume  of  the  cone. 

48.  The  altitude  to  the  base  of  an  isosceles  triangle  is 
5.4  feet,  and  one  of  the  equal  altitudes  is  7.12  feet.  Find 
the  area. 

49.  The  area  of  an  isosceles  triangle  is  equal  to  the  area 
of  the  semicircle  on  the  base  as  a  diameter.  Find  the  angle 
at  the  base. 

50.  In  a  circle  whose  radius  is  9.4  inches,  a  chord  of  14.4 
inches  is  drawn.  Find  the  central  angle  corresponding  to 
this  chord. 

51.  Calculate  the  area  of  a  circle  in  which  a  chord  5  feet 
long  subtends  a  central  angle  of  77°  36'. 

52.  The  lengths  of  the  sides  of  a  rectangle  are  78  ft.  and 
71  ft.  The  middle  points  of  its  sides  are  the  corners  of  an 
inscribed  rhombus.     Find  an  acute  angle  of  this  rhombus. 

53.  A  point  moves  in  the  arc  of  a  semicircle  of  radius  b 
with  a  uniform  velocity  of  a  feet  per  second.  In  how  many- 
seconds  is  it  at  a  distance  of  c  feet  (measured  on  the  chord) 
from  its  starting  point  ?     Take  b  =  10,  a  =  3,  c  =  16. 

54.  Two  equal  forces  act  on  a  body,  the  angle  between 
them  being  146°  48'.  Their  resultant  is  12  pounds.  Find 
the  component  forces. 

55.  What  must  be  the  angle  between  two  equal  forces, 
each  3.5  pounds,  if  their  resultant  is  1  pound  ? 


REGULAR  POLYGONS  AND   STARS. 


63 


THE  REGULAR  POLYGON. 

33.  Denote   the  number  of 

sides    by   n.      Then   the   half 

180° 

anerle   at   the  center  is   , 

°  n 

which  denote  by  A.  Also  de- 
note the  side  by  a,  the  radius 
of  the  circumcircle  by  E,  and 
the  radius  of  the  incircle  by  r. 
Then 

a  =  2  R  sin  A  =  2  r  tsin  A. 

R  =  rseGA  =  iacscA. 

r  =  ^a  cot  A  =  E  cos  A. 

Area  polygon  =  ^  nar  =  J  na^  cot  A. 

Area  incircle  =  v)-^  =  \  ircv^  cot^  A. 
\  ttci^  csc^  a. 


Fio.  36. 


Area  circumcircle  =  irE^ 


Prove  the  following  properties  of  regular  polygons : 

The  area  of  the  ring  between  the  incircle  and  circumcircle 
of  a  regular  polygon  is  equal  to  the  area  of  the  circle  on 
one  side  as  a  diameter. 

The  perimeter  of  a  regular  polygon  :  the  perimeter  of  its 
incircle  =  area  of  polygon  :  area  of  incircle. 

The  area  of  a  regular  inscribed  polygon  of  an  even  num- 
ber of  sides  is  a  mean  proportional  between  the  areas  of  the 
regular  inscribed  polygon  and  the  regular  circumscribed 
polygon  of  half  the  number  of  sides. 

Stars.  —  Divide  the  circumference  of  a  circle  into  n  equal 
parts.  Let  s  denote  a  whole  number  less  than  ^n-,  join 
every  sth  point  of  division  of  the  circumference  (or  every 
n  —  sth  point).  There  is  obtained  an  w-pointed  star  of 
species  s. 


64   ISOSCELES  TRIANGLE  AND  REGULAR  POLYGON. 


1.  If  s  is  1,  the  perimeter  is  continuous  and  non-inter- 
secting ;  the  figure  consists  of  a  regular  polygon  of  n  sides. 

2.  If  n  is  a  multiple  of  s,  say  n  =  rs,  the  perimeter  is 
discontinuous ;  the  figure  consists  of  s  regular  polygons 
each  of  r  sides.  For  example,  the  fifteen-pointed  star  of 
species  3  consists  of  3  regular  pentagons. 

3.  If  n  and  s  have  a  common  divisor,  say  n  =  ru  and 
5  =  rv,  the  perimeter  is  discontinuous ;  the  figure  consists 
of  r  different  stars  each  of  them  having  u  corners.  For 
example,  the  fifteen-pointed  star  of  species  6  consists  of  3 
five-pointed  stars. 

4.  If  n  and  s  are  prime  to  each  other,  the  perimeter  is 
continuous  and  intersecting.  For  example,  the  fifteen- 
pointed  star  of  species  4. 

Properties.  —  1.    In  a  star,  prove  that  the  sum  of  the 

71  angles  =  (n- 2  s)  180°. 

2.   Prove  that    a  =  2  i?  sin-  180°  =  2  r  tan-  180°, 

71  n 

R  being  the  radius  of  the  circumcircle,  r  the  radius  of  tlie 
incircle,  and  a  the  length  between  any  two  points  joined  in 
the  construction. 

Exercises.  —  The  following  table  gives  elements  of  seven 
regular  polygons.  Assume  the  number  of  sides  and  any 
one  element  to  compute  the  other  elements. 


n 

a 

B 

r 

K 

5 

24.68 

21.00 

16.99 

1049 

6 

20.78 

20.78 

18.00 

1122 

7 

13.88 

16.00 

14.41 

700 

8 

33.14 

43.30 

40.00 

5300 

12 

14.00 

27.00 

26.12 

2195 

20 

6.257 

20.00 

19.75 

1236 

25 

10.90 

43.50 

43.16 

5880 

APPLICATIONS.  65 

PROBLEMS. 

34.  56.  The  diagonal  of  a  regular  pentagon  is  4.3  feet  (d). 
Find  the  area  of  the  pentagon. 

57.  The  area  of  a  circle  is  15  sq.  feet.  Find  the  area  of 
a  regular  heptagon  whose  perimeter  is  equal  to  the  circum- 
ference of  the  given  circle. 

58.  Find  the  largest  and  smallest  diagonals  of  a  regular 
polygon  of  7  (2  n  -f  1)  sides,  each  side  being  4  yards  (a). 

59.  In  a  park  it  is  desired  to  make  a  flower  garden  to 
contain  one  acre,  and  to  be  of  the  form  of  a  regular  five- 
pointed  star.     Find  each  side  of  the  bounding  decagon. 

60.  Kegular  polygons  of  27  sides  are  inscribed  in  and 
circumscribed  about  a  circle  whose  radius  is  9.2,  the  sides 
of  the  two  polygons  being  respectively  parallel.  What  is 
the  area  of  the  figure  included  between  the  perimeters  of 
the  two  polygons  ? 

61.  The  area  of  a  regular  inscribed  polygon  of  n  sides 
is  K.  Find  the  area  of  a  regular  polygon  of  m  sides 
circumscribed  about  the  same  circle.  Put  n  =  9,m  =  7, 
K=  100. 

62.  In  a  circle  whose  radius  is  2  feet  two  parallel  chords 
are  drawn  on  the  same  side  of  the  center.  One  is  equal  to 
the  side  of  the  regular  inscribed  polygon  of  9  sides,  the 
other  to  the  side  of  the  regular  circumscribed  hexagon 
(about  the  same  circle).  Calculate  the  area  of  the  part 
included  between  these  two  chords. 

63.  The  radius  of  a  circle  is  7  yards.  A  chord  is  drawn 
equal  to  the  side  of  the  regular  inscribed  polygon  of  15 
sides.  In  the  larger  of  the  two  segments  thus  formed,  a 
circle  is  inscribed  touching  the  arc  in  its  middle  point. 
Calculate  the  area  of  a  regular  polygon  of  9  sides  circum- 
scribed about  the  latter  circle. 


CHAPTER   V. 

OBLIQUE  TRIANGLES. 


Fig.  37. 


Fig.  38. 


35.  Notation.  —  In  an  oblique  triangle,  let  A  denote  the 
value  of  angle  A,  B  the  value  of  angle  B,  C  the  value  of 
angle  C ;  a  the  length  of  the  side  opposite  A,  b  the  length 
of  the  side  opposite  B,  c  the  length  of  the  side  opposite  (7; 
p  the  perpendicular  from  C  on  c,  Cj  the  distance  from  A  to 
the  foot  of  the  perpendicular,  and  Cg  the  distance  from  B 
to  the  foot  of  the  perpendicular.  The  length  Ci  will  be 
regarded  as  positive  when  drawn  from  A  toward  B,  and 
negative  when  drawn  in  the  opposite  direction ;  Cg  will  be 
regarded  as  positive  when  drawn  from  B  toward  A,  and 
negative  when  drawn  in  the  opposite  direction. 

Relation  between  the  angles.  —  From  geometry, 

A  +  B-\-C  =  180°. 
Hence,  when  any  two  angles  of  a  triangle  are  given,  the 
third  angle  can  readily  be  found. 

36.  Methods  of  Solution. — At  least  three  independent 
conditions  must  be  given  to  make  the  solution  of  an  oblique 
triangle  determinate.  Suppose  that  these  three  conditions 
are  the  values  of  sides  or  angles.  When  possible,  draw  an 
altitude  such  that  two  of  the  given  parts  are  in  one  triangle. 

66 


METHODS  OF  SOLUTION.  67 

Now  find  a  value  of  this  altitude  in  terms  of  two  known 
parts  in  the  one  right  triangle.  Also  in  the  other  right 
triangle  find  a  value  of  this  same  altitude  in  terms  of  the 
other  given  part  and  a  desired  part.  These  values  of 
the  altitude  give  an  equation  in  which  the  desired  part  is 
the  only  unknown.     Thus, 

Given  one  side  and  two  angles :  as  6,  ^  and  B.  To  find  a, 
equate  values  of  p  thus :  p  =  asm  B  =  h  sin  A. 

Multiply  by  esc  B:  a  =  b  esc  B  sin  A. 

If  C  and  c  are  also  desired :    C  =  180°  -(A-\-  B). 
By  drawing  the  altitude  from  A  on  a, 

c  sin  2^  =  6  sin  C,  whence  c  =  b  esc  B  sin  C. 
These  values  of  a  and  c  give  the  following  relation : 

acscA  =  bcscB  =  c  esc  C.  (19) 

Given  two  sides  and  the  angle  opposite  one  of  these  sides: 
as  a,  b,  and  A.     Equate  values  of  p :  a  s'm  B  =  b  sin  A; 

whence  sin  B  =  -  sin  A, 

a 

C  and  c  being  found  as  in  the  last  case. 

Given  two  sides  and  the  included  angle :  as  b,  c,  and  A. 
First  find  Ci=  b  cos  A  and  Cj  =  c  —  q. 
To  find  B,  equate  values  of  p  thus : 

p  =  C2  tan  B=  Ci  tan  A ;  whence  tan  B  =  —  tan  A. 

To  find  a,  equate  values  of  p^ : 

a^  —  c^  =  6^  —  Ci.  Put  c  —  Ci  for  Cg. 

a2  =  52  +  (c  _  c,)2  _  ci^  =  62  _^  c2  -  2  (x^. 
Put  b  cos  A  for  Ci ;  then  a*  =  6^  +  c^  -  2  6c  cos  A.        (20) 
Given  the  three  sides :  as  a,  6,  and  c. 

From  (20),  cos  A  =  ^' '^/~ ''\  (21) 

with  similar  relations  for  the  other  angles. 


68 


OBLIQUE   TRIANGLES. 


It  is  usually  inconvenient  to  compute  cos  A  from  (21) ; 
another  method  will  now  be  found  for  computing  the  values 
of  the  angles  of  a  triangle  when  the  three  sides  are  given. 


Fig.  40. 


Prolong  the  sides  AB  and  ^C  of  triangle  ABC.  Bisect 
angles  A,  B,  and  GBC^,  these  bisectors  meeting  at  O  and  Ox. 
The  point  0  is  the  center  of  the  incircle  which  touches  the 
sides  of  the  triangle  in  Aq,  Bq,  and  Co ;  the  point  Oi  is  the 
center  of  the  excircle  in  angle  A  which  touches  BC  at  Aiy 
AC  prolonged  at  B^,  and  AB  prolonged  at  Ci. 

From  geometry,  the  tangents  from  a  point  outside  of  a 
circle  are  equal  to  each  other ;  thus  ABq  =  ACq,  ABy  =  AC^ 
and  so  on. 


Denote  each  of  the  tan- 
gents from  A  to  the  incircle 
by  X,  each  of  the  tangents 
from  B  by  y,  each  of  the 
tangents  from  C  by  z,  and 
half  of  the  sum  of  the  sides, 
|-  (a  4-  6  +  c)  by  s ;  then 


Denote  each  of  the  tan- 
gents from  A  to  the  excircle 
by  Xi,  each  of  the  tangents 
from  B  by  y^,  each  of  the 
tangents  from  G  by  z^,  and 
half  of  the  sum  of  the  sides 
|(a  +  6  +  c)  by  s ;  then 


THE  INCIRCLE  AND  EXCIRCLE. 


69 


2^  4-  2  =  a, 

which  give 

X  =  ABq  =  ACq  =  s  —  \i, 
y  =  BCo  =  BAo=8-b, 
z  =  CAq  =  CBq  =  s  —  c. 


yi  +  Zi  =  a, 

Xi-Zi  =  b, 

«i  -  2/1  =  c, 
which  give 
Xi  =  ABi  =  ACi  =  s, 
y,  =  BCi  =  BA^  =  s-c, 
z,  =  CA,  =  CB,=s-b. 


Now  the  two  triangles  AOCq  and  AOiCi  are  similar  be- 
cause their  sides  are  respectively  parallel;  and  the  two 
triangles  BOCq  and  BOiCi  are  similar  because  their  sides 
are  respectively  perpendicular ;  hence 


ACo 


or 

r      s  —  a 
r,         s 

or   - 

r         s  —  b . 

ACi 

OCq^BCq 
BCi 

r  and  rj  being  the  radii  of  the  incircle  and  excircle  respec- 
tively. 
Multiply  these  equations  and  solve  for  r. 


In  triangle  A00„,      cot  ^  j«  =  ?^^ 

P 
A 

Changing  letters, 


cot^B  = 


cot^C 


r 

—  c 


(22) 


The  equations  in  (22)  furnish  the  solution  for  the  angles 
of  a  triangle  when  the  sides  are  given.  This  method  is 
more  suitable  for  logarithmic  computation  than  the  use 
of  (21). 


70  OBLIQUE   TRIANGLES. 


SOLUTIONS. 


One  side  and  two  angles. 

37.   1.  Given  a  =  200,  A  =  74°  36',  B  =  81°  12' ;  find  b. 

By  (19),     b  =  acsGAsmB^  200  x  1.0372  x  .9882  =  205. 

2.  With  the  same  data,  find  C  and  c. 

C=  180°  -(A  +  B)  =  180°  -  155°  48'  =  24°  12'. 
By  (19),  c  =  acscA  sin  (7=200  x  1.0372  x  .4099  =  85. 

3.  Given  &  =  37,  5  =  53°  8',  ^  =  107°  57';   find  a. 

By  (19),     a  =  6  CSC  ^  sin  ^  =  37  x  1.2500  x  .9513  =  44. 

Two  sides  and  the  angle  opposite  one  of  them. 

1.  Given  A  =  30°,  a  =  1,  ft  =  4 ;   find  5  and  C. 
Here  sin5  =  f  sin30°  =  4  x  .5  =  2. 
Since  sin  B>1,  the  solution  is  impossible. 

2.  Given^  =  30°,  a  =  2,  6  =  4;   find  ^  and  C. 
Here  sin  5  =  f  sin  30°  =  1. 
Whence   B  =  90°  and  C  =  180°  -(A-\-B)  =  60°. 

3.  Given^  =  30°,  a  =  3,  6  =  4;   find  ^  and  C. 
Here  sin  ^  =  J  sin  30°  =  .6667. 

B  =  41°  49'  or  138°  11',  and  C  =  108°  11'  or  11°  49'. 
Note  that  the  values  of  C  are  obtainable  by  subtracting 
A  from  the  values  of  B.     Explain  this  geometrically. 

4.  Given  A  =  30°,  a  =  4,  6  =  4 ;   find  5  and  C. 
Here  sin  ^  =  J  sin  30°  =  .5. 

B  =  30°  or  150°,  and  C  =  120°  or  0°. 

5.  Givenyl  =  30°,  a  =  5,  6  =  4;   find  5  and  a 


Here  sin  B  =  ^sm  30°  =  .4. 

B  =  23°  35'  or  156°  25',  and  C  =  126°  25'  or  -  6°  25 


SOLUTIONS  USING  NUMBERS.  71 

These  five  examples  are  illustrated  in  the  geometric  con- 
struction, Fig.  41.  The  base  is  drawn  indefinitely,  angle  A 
is  made  equal  to  30°,  AC  =6  =  4.  Successive  radii,  Oj,  a^, 
ttg,  a^,  Qs  are  used  to  draw  arcs  with  the  center  at  C. 

In  order  that  the  triangles  may  have  the  angle  A  =  30°, 
the  vertex  B  of  each  of  them  must  be  on  the  half -line  ABa 
drawn  through  A  to  the  right.  Also  in  order  that  the  tri- 
angles may  have  the  side  a  of  the  length  given  in  each 
problem,  the  vertex  B  must  be  in  the  arc  appropriate  to 
that  problem. 

To  prevent  confusion  in  the  diagram,  the  side  a  has  not 
been  drawn,  except  for  the  negative  solution  of  question  5. 
It  is  recommended  that  the  boundary  of  the  triangles  be 
drawn  with  crayons  of  different  colors. 

C 

"    /'    /'     ''    «. 


bT^  7S      /'b,      /b, 


Fig.  41. 

It  will  be  noticed  that  the  first  arc  does  not  intersect  the 
half -line  AB^.  The  second  arc  touches  it  at  one  point.  The 
third,  arc  cuts  it  at  two  points,  both  to  the  right  of  A.  The 
fourth  arc  cuts  it  once  to  the  right  of  A  and  once  at  A. 
The  fifth  arc  cuts  the  half-line  AB^  to  the  right  of  A  and 
also  cuts  the  half-line  AB^  to  the  left  of  A\  the  negative 
value  of  0=:^-^°  25',  being  the  angle  ACBi, 


72 


OBLIQUE  TRIANGLES. 


Solve  the  following  five  examples,  and  compare  the  results 
with  the  geometric  construction,  Fig.  42. 

1.  Given  ^  =  150°,  a  =  l,  6  =  4;    find  5  and  C. 

2.  Given  ^  =  150°,  a  =  2,  5  =  4; 


3.  Given  ^  =  150°,  a  =  3,  &  =  4; 

4.  Given  ^  =  150°,   a  =  4,  6  =  4; 

5.  Given  A  =  150°,  a  =  5,  6  =  4; 

C 


find  5  and  C. 
find  5  and  C. 
find  B  and  (7. 
find  B  and  a 


■A"'bV\'""bJ\T         " 


\. 


Fig.  42, 


Assume  a  value  in  each  of  three  columns  in  any  row  in 
the  following  table,  and  compute  the  remaining  value  (or 
values)  in  that  row : 


a 

h 

^ 

B 

14 

15 

59°  29' 

67°  23'  or  112°  37' 

37 

40 

67°  23' 

93°  42'  or    86°  18' 

75 

77 

61°  56' 

64°  57'  or  115°   3' 

69 

139 

21°  14' 

46°  52'  or  133°   8' 

80 

401 

11° 27' 

84°  17 'or    95°  43' 

101 

120 

43°  36' 

55°    1'  or  124°  59' 

148 

153 

53°   8' 

55°  48'  or  124°  12' 

240 

409 

34°   7' 

72°  56'  or  107°    4' 

449 

560 

51°  25' 

77°  10'  or  102°  50' 

SOLUTIONS  USING  NUMBERS. 


73 


6.  Given  a  =.=  7,  6  =  8,  A  =  60°)   find  c  directly. 
From  (20),  d" -2  be  cos  A  =  a"  -  h\ 
Substituting  the  given  numerical  values, 

c^  -  8  c  =  -  15, 
which  gives  c  =  3  or  5. 

7.  Given  a  =  7,  6  =  5,  ^  =  120°;    find  c. 
Here  the  equation  is  c^  +  5  c  =  24. 
Therefore  c  =  3   or    —  8. 

Two  sides  and  the  included  angle. 

1.  Given  6  =  85,  c  =  200,  ^  =  8ri2';    find  ^. 
Ci  =  6  cos  ^  =  85  cos  81°  12'  =  13. 
C2  =  200-13  =  187. 
p  =  Cg  tan  B  =  Ci  tan  A. 

tan  J5  =  ^  tan  ^  =  .4491 
Therefore       B  =  24°  11. 

2.  With  the  same  data,  find  a. 

From  (20),  a-  =  6^  +  c^  -  2  6c  cos  ^ 

=  7225  +  40000  -  5200  =  42025. 
Therefore  a  =  205. 

3.  Given  6  =  37,  c  =  15,  ^  =  107°  57';    find  5. 
Here     Ci  =  6  cos  A=  —  11.4. 

C2=  15 -(-11.4)=  26.4. 

tan  5  =  ^  tan  ^  =  - 1^4  X  (- ^-^868) 
C2  26.4 

=  1.3329.    Therefore,  B  =  53°  7'. 

4.  With  the  same  data,  find  a. 
a'  =  b^-\-€^-2bccosA  =  1369  +  225  +  342 

Therefore,  a  =  44. 


74  OBLIQUE  TRIANGLES. 

The  three  sides. 

1.    Given  a  — 7,  b  =  8,  c  =  5;    find  A. 

2  be  2x8x5 

Or  the  angle  may  be  found  by  using  the  first  and  second 
equations  in  (22) ;  thus, 


2.    Given  a  =  7,  6  =  5,  c  =  3 ;    find  J.. 

cos  A  =  ^'+/-^'  =  _  .5.     Therefore  A  =  120°. 

2  be 


cot  -^ ^  =-J; '"  ^  :"^  =  .5774 ;   1^  =  60° 


7.5  X  .5 
2""      \2.5x4.5 


Exercises. —  Assume  any  three  parts  in  each  of  the  fol- 
lowing five  triangles.     Then  compute  the  fourth  part : 

^  =  60°  a  =  13  &  =  15  c  =  7Gr8 

A  =  120°  a  =  13  6=7  c  =  8or-15 

^  =  53°8'  a  =  13  6  =  15  c  =  4orl4 

^  =  61°  56'  a  =  65  6  =  68  c  =  7or57 

^  =  67°  23'  a  =  25  6=26  c  =  3  or  17 

38.  Given  two  sides  and  the  included  angle,  to  find  a  func- 
tion of  another  angle  directly  in  terms  of  the  given  parts. 

Since  c  =  c^  -f  Cg : 

c=p  (cot  A  +  cot  B). 

Multiply  by  esc  A,  and  put  6  for  ^  esc  ^ : 

c  CSC  ^  =  6  (cot  A  +  cot  B).  (23) 

Whence  cotB  =  c^-^-cotA.  (24) 

0 

CSC  A. 
After  B  has  been  found,  a  =  GscB-. — - — 


BROCARD'S  POINTS.  75 

The  objection  to  the  use  of  (24)  is  that  in  general  both 
tables  of  functions   are   used;   the  table  of  logarithms  of 

functions  in  the  computation  of  the  value  of  c  ^^^     ,  and 

b 
natural  values  of  functions  in  completing  the  work.      In 
general,  this  computation  is  not  troublesome;   and  where 
results  are  to  be  carried  through  a  series  of  triangles,  as  in 
a  survey,  it  is  of  advantage  to  be 
able  to  write  down  the  function  ^,^ 

of   a  desired   angle   directly  in         /*  >^\ 
terms  of  the  data.  /  X  ^^"\"\ 


Fia.  45. 


As  a  simple  application  of  (23),    Jc- 
consider  the  following  problem  : 

When  a  point  (X)  in  a  triangle 
is  joined  to  the  corners,  it  is  found  that  the  angles  XBA^ 
XCB,  and  XAO  are  all  equal.  Find  the  value  of  one  of 
these  equal  angles  (jr). 

In  triangle  AXC,  by  (23) ;  ^Xcsca;=6(cota;+cot  CXA). 

In  triangle  AXB,  by  (19) ;  AX cscx  =  c esc  AXB. 

Now  angle  CXA  =  180°  -(C-x)-x  =  180°  -  C. 

Similarly,  AXB  =  180°  -  A. 

Hence  by  substituting  :  b  (cot  x  —  cot  C)  =  c  esc  A. 

But  by  (23),   c  esc  A  =  b  (cot  A  -h  cot  B).    Hence 

cot  X  =  cot  ^  +  cot  -B  4-  cot  C. 

If  the  angles  are  65°,  30°,  and  85°,  respectively,  x  =  23°  38'. 

The  point  X  and  another  point  found  by  counting  the 
angles  x  from  the  other  sides  of  a,  b,  and  c  are  called 
Brocard's  Points.  They  possess  many  interesting  properties; 
for  example,  they  are  the  foci  of  the  ellipse  inscribed  in 
the  triangle. 


76  OBLIQUE  TRIANGLES. 

THE  AREA   OF  A  TRIANGLE. 
39.   Denote  the  area  by  K.     Then 
By  geometry,  K=\cp. 

Since  ^  =  6  sin  ^,  -K"  =  ^  he  sin  A. 

Changing  letters,  K=^ca  sin B. 

K=^ab  sin  C. 
b  =  acsGAsmB,  K=  ^a^  csGAsinBsinC.    ' 

Changing  letters,  ^=  ^  6^  sin  ^  esc  B  sin  C. 

^=  i  c^  sin  J.  sin  5  CSC  (7. 
From  Fig.  40,  K=^ar  +ibr  +  ^ cr  =  rs. 

Substitute  from  (22),    K  =  Vs{s -a)(s -b)(s- c). 


Prove  the  following  :       K=  \-wa?bh^  sin  A  sin  B  sin  G. 

K=i^Qo\,\Acoi\BGoi^C. 

7ir=  ri^  cot  i  ^  tan  1  ^  tan  1  O. 

K=rriCot^A. 

Kz=  ^  tan  \  A  tan  \  B  tan  \  C. 

/ir=  (s  -  a)nan  1  ^  cot  i  J5  cot -J  C. 

By  equating  any  two  values  of  the  area,  a  relation  between 
segments  and  angles  of  the  triangle  is  obtained. 

Exercises. — Without  using  logarithms,  find  the  area  of 
each  of  the  following  triangles : 

a  =  115  ^  =  67°  23'  (7  =46°  24' 

a  =  240  6  =  53  B=    8°  10' 

a  =  200  6  =  205  C=24°ll' 

a  =  116  6  =  105  c  =  143 


SOLUTIONS  USING  LOGARITHMS.  77 

LOGARITHMIC   SOLUTIONS. 
40.   Given  one  side  and  two  angles.  — 

a  =  24.31,   ^  =  45°  18',   5  =  22°  11'. 
Solve  the  triangle. 

"v-    =  a  CSC  ^  sin  5,   c  —  a  esc  A  sin  C 


isinS     9.5770 

log  a         1.3858 

+  ^=   67°  29' 

log  CSC  ^  0.1483 

(7  =112°  31' 

L  sin  C     9.9656 

h=    12.92 

log  6         1.1111 

c=   31.6 

logc         1.4997 

Given  two  sides  and  the  angle  opposite  one  of  them. — 
a  =  215.9,  h  =  307.7,  A  =  25°  10'. 

Solve  the  triangle, 
sin  5=  -  sin  ^,  Ci  —  Bi  —  Af    C2  =  Bi  —  A,   c  =  a  csc  J.  sin  C. 

(X 


log  6 
log  a 

LsinA 

LsinB 

L  sin  d 
log  a 
log  CSC  A 
L  sin  C2 

logCi 

l0gC2 

2.4881 
2.3343 

0.1538 
9.6286 

JBi=    37°  18' 

9.7824 

^2  =  142°  42' 
A=   25°  10' 
Ci  =  117°  32' 
C2=   12°   8' 

9.9478 

2.3343 

L  0.3714 

9.3226 

ci  =  450.3 
C2  =  109.2 

2.6535 
2.0383 

78 


OBLIQUE   TRIANGLES. 


Given  two  sides  and  the  included  angle. — 
h  =  767,  c  =  947,  A  =  10°  50'. 
Find  the  remaining  parts. 

Ci  =  &  cos  A,  Co=  c  —  Cj, 

p  =  G2  tan  B  =  Ci  tan  A  j 

whence  tan  B  =  —  tan  A,  a  =  C2  sec  B. 
C2 


c  =  947 
Ci  =  753.3 
C2  =  193.7 

A=    10°  50' 
^=    36°  40' 


47°  30' 
0=132°  30' 
a  =  241.5 


log  6 
LcosA 
logci 
logca 

LtSinA 
LtSinB 
log  sec  B 

log  a 


2.8848 
9.9922 
2.8770 
2.2871 
0.5899 
9.2819 
9.8718 
0.0958 
2.2871 


2.3829 

Given  b  =  767,  c  =  947,  A  =  10°  50' ;  find  a. 
By  (20), 


a  =  V6^  +  c^  —  2  6c  cos  ^. 


2  6c  cos  JL 


:  588300 
:  896600 
1484900 
1427000 
a'  =  57900 
a  =      240.6 


log  2 
log  6 
logc 
icos^ 
log  62 

logc^ 


0.3010 
2.8848 
2.9763 
9.9922 


5.7696 
5.9526 
log  2  6c  cos  ^  6.1543 
loga^  4.7627 

log  a  2.3813 


Two  significant  figures  were  lost  in  the  subtraction  of 
2  6c  cos  A  from  6^  +  c-.  Hence  the  lack  of  precision  in  the 
value  of  a. 


SOLUTIONS  USING  LOGARITHMS. 


79 


Given  h  =  .049,  c  =  .045,  A  =  143° ; 
find  the  remaining  parts. 


I  =  b  cos  A,  C2—  c  — 

•Ci, 

\  X 

tan  5  =  -  tan  A, 

C'2 

a  =  Cg  sec  B. 

L   .^'   \a   c  \ 

Fig.  47. 

c  =      .045 

Lb 
LcosA 

Lci 

8.6902 
9.9023  n 

Ci  =  -  .03913 

8.5925  w 

C2  =      .08413 

Xtan^ 
L  tan  5 
log  sec  B 

LC2 

La 

8.9250 

A  =  143° 

9.6675  n 
9.8771  n 

J5=    19°  19' 

9.5446 

162° 19' 
C=    17°41' 

0.0252 
8.9250 

a  =.08917     . 

8.9502 

Given  b  =  .049,  c  =  .045,  A  =  143° ;  find  a. 


a  =  V62H-c2 

—  2  be  cos 

A. 

log  2 

0.3010 

Lb 

8.6902 

Lc 

8.6532 

LeosA 

Lb' 

9.9023  n 

.002401 

7.3804 

.002025 

Lc' 

7.3064 

.003521 

L2be  cos 
La' 

iA  7.5467  n 

.007947 

7.9002 

.08915 

La 

8.9501 

80 


OBLIQUE   TRIANGLES. 


Given  the  three  sides.  —  a  =  704, 
Find  the  angles  and  the  area. 


6  =  302.  c  =  670. 


4 


Ks-a){s- 
s 

b)(s- 

■""^  coti^  =  ^~ 

a 

—,  etc. ; 

704 

134 

logs  — a 

2.1271 

302 

536 

log  s  —  6 

2.7292 

670 

168 

log  s  —  c 

2.2253 

1676 

7.0816 

868 

868 

logs 

log  7^ 

2.9232 

4.1584 

logr 

2.0792 

83°  42' 

log  cot  ^  A 

0.0479 

25° 14' 

log  cot  ^  B 

0.6500 

71°   4' 

log  cot  ^  C 

0.1461 

100,550 

logK 

5.0024 

/r=  rs. 


Exercises.  — Assume  three  values  in  any  row  of  the  follow- 
ing table  and  compute  the  remaining  values  in  that  row. 


a 

h 

c 

A 

B 

K 

6.82 

5.20 

3.16 

106°  47' 

46°  53' 

7.864 

.317 

.533 

.510 

35°  18' 

76° 19' 

.07854 

28.9 

60.1 

71.2 

23°  32' 

56°   9' 

854.3 

1.98 

2.02 

3.16 

37°  22' 

38°  16' 

1.937 

312 

109 

229 

131°  25' 

15° 11' 

9360 

14.8 

17.5 

15.3 

53°   8' 

71°   5' 

107.1 

1.75 

1.62 

1.19 

67°  23' 

73°  44' 

99.96 

145 

119 

156 

61°  56' 

46°  24' 

8190 

11.6 

40.4 

48.0 

11°  25' 

43°  36' 

192.0 

SOLUTIONS  OP  PROBLEMS. 


81 


41.  Whenever  it  is  possible,  by  means  of  the  ruler  and 
compasses,  to  obtain  the  geometric  construction  of  a  prob- 
lem, the  trigonometric  solution  is  always  obtainable  by  follow- 
ing the  order  of  the  construction.  It  occasionally  happens, 
however,  that  the  geometric  construction  is  not  possible. 
When  this  occurs,  the  trigonometric  solution  produces  an 
equation  of  a  higher  degree  than  the  second.  In  many  such 
cases,  the  method  of  solution  by  trial  involves  the  least  labor. 

As  an  illustration,  consider  the  following  problem : 

A  ship  is  sailing  in  the  direc- 
tion N.  35°  W.  When  at  B,  two 
lighthouses  are  sighted,  one  due 
north,  the  other  due  west.  After 
sailing  3  miles,  the  ship  was 
equally  distant  from  the  light- 
houses; sailing  1  mile  further 
on  the  same  course,  the  ship  and 
the  lighthouses  are  in  the  same 
straight  line.  Find  AC^  the  dis- 
tance between  the  lights. 

From  D,  the  second  position 
of  the  ship,  drop  a  perpendicular 
to  F,  the  middle  of  AC.  Denote  angle  A  by  a* ;  then  the 
geometry  of  the  figure  gives:  ?i\\^\e  BFE  =  2 x,  BEF= 
125° -a;,  BFD  =  2x-^0°,  and  BDF  =  215° -x. 

Apply  (19)  to  triangles  BFD  and  BFE : 
BF=  3  esc  (2  a;  -  90°)  sin  (215°  -x)  =  4.csc2x  sin  (125°  -  x), 
or     -*  3  sec  2  a;  cos  (125°  -  x)  =  4  esc  2  a;  sin  (125°  -  x). 

Multiply  by  sin  2  a;  sec  (125°— ar)  and  transpose  : 
—  3  tan2a;- 4  tan(125°  -  a;)  =  0. 

Now  solve  this  by  trial.     Substituting  0°  for  x  makes  the 
left-hand  member  of   the  equation   positive,  90°—,  45°+, 
60°-,  50°+,  52°-,  and  so  on;  a;  =  51°  7'.     Hence 
AC=2BF=S  esc  102°  14'  sin  73°  53'  =  7.864  miles. 


Fig.  48. 


82 


OBLIQUE    TRIANGLES. 


In  order  to  measure  the  dis- 
tance between  two  inaccessible 
points,  C  and  D,  a  base  line 
AB  was  laid  off  equal  to  583 
feet  (a).  At  A  angle  CAD 
=  34°  and  angle  DAB  =  41° 
were  measured.  Also  at  B,  the 
angle    ABC  =  44°    and    angle        ^ 

CBD  =  32°.     Find  CD.  ^'  I       ; 

Fig.  49. 

^(7=acsc^(7i?sin^J5(7;  AD=acsGADBsmABD 

CD' = AC + AD'- -2  AC X  AD  COS  CAD 

L  sin  ABC     9.8418 


^B 


(25) 


ACB  = 

=  55° 

log  CSC  ACB 

0.0866 

log  a 

2.7657 

ADB  = 

=  57° 

log  CSC  ADB 

0.0764 

ABD  = 

=  70° 

L  sin  ABD 

9.9869 

log  AC 

2.6941 

AC'  = 

=  244500 

log  AD 

2.8290 

AD''-- 

=  455000 

log  2 

0.3010 

699500 

L  cos  CAD 

9.9186 

553000 

5.7427 

log  AC' 

5.3882 

log  AD' 

5.6580 

CD'-. 

=  146500 

log  CD' 

5.1658 

CD-- 

=  382.7 

log  CD 

2.5829 

In  making  an  extended  survey,  it  may  happen  that  CD, 
Fig.  49,  is  a  base  of  known  length,  the  angles  being  meas- 
ure(f  from  A  and  B.  In  this  case,  to  find  AB,  assume  the 
figure  drawn  to  scale,  AB  being  represented  by  1  foot.  In 
this  reduced  figure,  CD  may  be  computed ;  then  the  numeri- 
cal value  oi  AB  equals  the  ratio  of  the  known  length  of  CD 
to  the  computed  length  of  CD  in  the  reduced  figure. 


SOLUTIONS   OF    PKOBLKMS. 


The  sides  of  a  trian- 
gle are :  a  =  57,  b  =  51, 
c  =  64.  A.  point  is  taken 
in  the  side  c,  such  that 
the  segments  of  c  are : 
/i  =  21,  A;  =  43.  Find  I, 
the  distance  of  this  point 
from  C. 

Denote    the    angle   be-  ^ 
tween  h  and  I  by  D. 

By  (20),  l^  +  h^-2  hi  cos  D  =  b^', 

By  (20),      P-\-k--2M  cos  (180°  -  D)  =  a\ 
Multiply  the  first  by  k;  the  second  by  h,  and  add : 

cl'  -h  chk  =  a'h  +  b'k.  (26) 

Substituting,  I  is  found  to  be  43.71. 

The  top  of  a  lighthouse,  known  to  be  100  feet  (h)  high, 
is  just  seen  in  the  horizon.  What  is  the  distance  (6)  of  the 
lighthouse  from  the  observer  ? 

It  is  known  that  near  the  earth's  surface  light  does  not 
move  in  a  horizontal  line ;  on  account  of  refraction,  a  ray, 
which  seems  horizontal,  moves  in  a  curve,  which  is  nearly 
the  arc  of  a  circle  whose  radius  is  7  times  the  earth's  radius. 

In  (26)  put  E  for  h,  (y  R  for  k,  7  E  for  a  or  c,  E  +  h  for  I 
Then  6b^  =  T{2Eh-\-  1i'). 

In  the  problem  under  consideration  li^  is  very  small  com- 
pared to  2  Ell,  and  may  be  neglected.  Now  h  is  usually 
given  in  feet,  and  b  is  desired  in  miles.  Neglect  1i^  and 
multiply  the  left  side  of  the  equation  by  5280 : 

31680  b^  (miles)  =  14  E  x  (Ji  in  feet). 

Put  3960  for  E  and  solve  for  b. 


b  (in  miles)  =  1.323  V/i  (in  feet). 
Substituting,  b  =  13.23  miles. 


(27) 


84 


OBLIQUE   TRIANGLES. 


Fig.  51. 


Snell's  Problem.  — On  a  coast  are  three  prominent  objects, 
A,  C,  and  B.  CB  =-  1100  feet  (a),  ^IC-:  1800  feet  (b),  and 
the  angle  ACB  =  150°  (0).  A  buoy  is  placed  at  a  point 
Xy  and  from  a  boat  at  the  buoy  the  horizontal  angles 
AXC  =  80°  {D)  and  CXB  =  60°  {E)  are  measured.  Find 
the  distance  from  X  to  C* 

The  following  values  are  obtained  by  following  the  geo- 
metric construction :       angle  AOC  =  2  D  (same  arc), 
angle  BO^C  =2E,  angle  OCO,  =  C -{- D  +  E  -  180°. 

Hence      OC  =  \a  esc  D,  and  OiC  =  \h  q^g E. 

Now  denote  angle  COO^  =  XAC  by  A.     From  (24), 


COtyl= CSC 

b 


DsmEcsG(C-\-D-i-E)-cot(C+D+E)  1 


CX=  a  CSC  D  sin  A 


(28) 


*  By  many  writers  the  credit  has  been  given  to  Pothenot  for  the 
suggestion  of  this  method  of  determining  the  position  of  a  point  by 
means  of  angles  subtended  by  segments  joining  three  known  points. 
The  problem,  however,  was  proposed  and  solved  in  the  year  1617  by 
Willebrord  Snell,  the  discoverer  of  the  law  of  the  refraction  of  light. 
John  Collins  gave  a  solution  of  the  problem  in  the  Philosophical 
Transactions  for  1671  :  Pothenot's  solution  was  in  1692. 


SNELL'S   PROBLEM. 


85 


3.0125 
3.2553 


i'xtermsof  thedata,    CX=^^^^^^^' 

^ -  ab  CSC  D  CSC  E  /It^  [C-^jyi^s) 

iL/a^  csc2  D  4-  6^  csc^  ^  +  2  aft  esc  1)  esc  ^  cos  (O  +  Z>  +  jEJ) 
The  computation  by  (28)  proceeds  thus : 
log  (-  a)       3.0414  71  .5719     log  a  esc  D  3.0480 

log  esc  Z>  0.0066  nat  cot  290° -.3640  Xsin^  9.8634 
L^x^E  9.9375  natcot^  .9359  log  CX  2.9114 
log  esc  290°  0.0270  n        A  =  46°  54'  CX  =  815.5 

log  6  

9.7572 

In  laying  out  a  railroad 
along  the  seashore,  the  dis- 
tances GB  and  DE  were 
measured  (C,  B,  D,  and  E 
all  in  a  straight  line) ;  also 
the  angles  CAB,  BAD,  and 
DAE  were  measured.  It 
was  found  that  CB  =  2200 
(a),  DE  =  2300  (b),  CAB  = 
21°  30' (A),  5^Z>  =  20°26' 
(^2),  and  DAE  =  22°A0' 
(As)  ;    find  BD.  Fig.  52. 

Denote  BD  by  x.     Then 

In  triangle  ADC,  x  -\-a  =  AG  esc  D  sin  (Ai  -f  A2). 

In  triangle  AEB,  x  -\- b  =  AE  esc  B  sin  (^2  +  -^3)- 

In  triangle  ABC,      AC  =  a  esc  A^  sin  B. 

In  triangle  AED,      AE  =  6  esc  ^3  sin  D. 
The  product  of  these  four  equations  is 
(x -{-a)(x-\-b)= ab  esc  A^  esc ^3  sin (A^^  +  ^2)  sin (A^ + ^3) .  (29) 
This  quadratic  equation  gives  x  =  1800. 


86 


OBLIQUE   TRIANGLES. 


There  are  two  observatories  on  the  same  meridian :  A  in 
Sweden,  latitude  65° 30'  N.  (I) ;  B  at  the  Cape  of  Good  Hope, 


Fig   53. 


latitude  34°  20'  S.  (—  T).  When  the  moon  passed  this  merid- 
ian, the  angles  ZAP  =  62°  20'  (z)  and  Z,BP  =  39°  20'  (z,) 
were  measured.  The  radius  of  the  earth  being  3960  miles, 
find  the  distance  from  the  moon  to  the  earth's  center. 

The  angles  of  the  quadrilateral  AEBP  are  readily  found : 
at  E,  the  angle  is  l  —  lx;  at  A,  it  is  180°  —  2; ;  at  P,  it  is 
z  +  z^-l  +  l^.     Also  the  angle  ABP  is  90°  +  i  0  -^  -  2  z^). 

In  triangle  AEB,    AB  =  2AE  sin  i  AEB. 

In  triangle  ABP,    AP  =  AB  esc  APB  sin  ABP. 
Denote  AE  by  r,  AP  by  v,  EP  by  x,  angle  xiEP  by  w. 


v=2  r  sin  ^{l—lj)  esc  (z-^z^—l-^-li)  cos^ 

Q-h-2z,) 

cot  w  = 

r 

=  -  CSC  z  +  cot  z 

V 

(30) 

x  = 

=  v(iSQw  sin  z 

log  2                 0.3010 

L^            8.2257 

V 

log  CSC  2    0.0122 

8.2379 

.0173 

nat  cot  2      .2401 

nat  cot  ic    .2574 

logv         5.3720 

logr                  3.5977 
isin    49°  55'  9.8837 
log  CSC   1°30'  1.6332 
Lcos    25°  15'  9.9564 
logv                  5.3720 

log  CSC  w  0 

L  sin  z      9 

5 

a:  =  236,40 

.0139 

.9878 
.3737 
Omi. 

APPLICATIONS.  87 


PROBLEMS. 


42.   64.   Solve  Problem  10  by  logarithms. 

65.  Solve  Problem  11  by  logarithms. 

66.  Solve  Problem  12  by  logarithms. 

67.  Solve  Problem  13  hy  logarithms. 

68.  Solve  Problem  14  by  logarithms. 

69.  Solve  Problem  15  by  logarithms. 

70.  To  measure  the  distance  between  two  points  A  and  C 
(C  inaccessible),  a  base  line  AB  =  093  feet  was  measured 
from  A.  At  A,  the  angle  subtended  by  BC  was  68°  29' ; 
and  at  B,  the  angle  subtended  by  AC  was  66°  7'.  Find  the 
distance  AC. 

71.  To  find  the  distance  AB  across  a  pond,  a  third  point 
C  was  taken.  AC  =735  yards,  jB(7=840  yards,  and  the 
angle  ACB=  55°  40'  were  measured.     Fiml  AB. 

72.  From  a  window  on  a  level  with  the  bottom  of  a 
steeple,  the  .angle  of  elevation  of  the  top  is  40°.  From 
another  window  18  feet  directly  above  the  former  the 
angle  of  elevation  is  37°  30'.  Kequired  the  height  of  the 
steeple. 

73.  A  tower  50  feet  high  is  situated  on  a  cliff  over  the 
sea.  From  the  bottom  of  the  tower,  the  angle  of  depres- 
sion of  a  ship  at  anchor  is  found  to  be  25°  17' ;  and  from 
the  top  the  angle  is  31°  12'.  Find  the  height  of  the  cliff 
and  the  direct  distance  of  the  ship  from  the  top  of  the 
tower. 

74.  A  frigate  10  miles  S.W.  of  a  harbor  sees  a  ship  sail 
from  it  in  the  direction  S.  70°  E.  at  the  rate  of  9  miles 
an  hour.  In  what  direction  and  at  what  rate  must  the 
frigate  sail  in  order  to  come  up  with  the  ship  in  two 
hours  ? 


88  OBLIQUE   TRIANGLES. 

75.  A  person  is  at  the  bank  of  a  river  directly  opposite 
a  tower  on  the  other  side.  Going  backward  from  the  bank, 
up  a  slope  making  an  angle  of  42°  to  the  horizontal,  he 
measures  an  oblique  distance  of  528  feet.  He  there  finds 
the  angle  of  depression  of  the  top  of  the  tower  is  90°,  and 
of  the  base  is  27°.  Required  the  height  of  the  tower  and 
the  breadth  of  the  river. 

76.  To  measure  the  breadth  of  a  stream,  a  base  AB  was 
chosen  =  490  yards  long,  parallel  to  the  bank  and  50  yards 
from  it.  A  tree  C  stands  on  the  other  bank  of  the  river. 
The  angle  CAB  wsiS  found  to  be  62°  37'  and  CBAA0°2S\ 
What  was  the  breadth  of  the  stream  ? 

77.  At  the  foot  of  a  mountain  the  elevation  of  the  sum- 
mit =  32°  57'.  A  person  measures  775.3  feet  along  the  slope 
toward  the  summit,  making  an  angle  of  7°  15'  with  the  hori- 
zontal plane ;  at  this  point,  the  elevation  of  the  summit  was 
60°  32'.     Find  the  height  of  the  mountain. 

78.  A  tower  stands  on  a  hill  whose  slope  is  uniformly 
26°  37'.  An  observer  at  the  foot  of  the  hill  finds  that  the 
tower  subtends  an  angle  of  10°  22'.  Measuring  horizontally 
up  the  slope  a  distance  of  200  feet,  he  observes  the  tower 
subtends  an  angle  of  20°  30'.     Find  the  height  of  the  tower. 

79.  A  tower  is  situated  on  a  hill.  At  a  point  in  the  hori- 
zontal plane  through  the  foot  of  the  hill,  the  angle  of  elevar 
tion  of  the  top  of  the  hill  is  40°  and  of  the  top  of  the  tower 
is  51°.  Measuring  in  a  line  directly  away  from  the  hill 
a  distance  of  180  feet,  the  angle  to  the  top  of  the  tower  is 
found  to  be  33°  45'.     Find  the  height  of  the  tower. 

80.  From  a  hill  200  feet  above  the  sea,  the  top  of  a  mast, 
known  to  be  150  feet  above  water,  was  just  seen  in  the 
horizon.     How  far  distant  is  it  ? 

81.  A  body  is  thrown  with  a  velocity  of  25  feet  per 
second  horizontally  from  the  window  of  a  railway  carriage 
moving  at  the  rate  of  30  miles  an  hour;   the  direction  in 


APPLICATIONS.  89 

which  the  body  is  thrown  makes  an  angle  of  30°  with  the 
rear  of  the  train.  Find  the  direction  of  the  vertical  plane 
in  which  the  body  moves. 

82.  A  man  traveling  at  the  rate  of  6  miles  an  hour  on 
a  road  that  went  due  east  observed  that  the  wind  struck 
him  from  the  northeast.  But  having  occasion  to  stop,  he 
found  that  it  actually  came  from  the  direction  N.  35°  E. 
Find  the  velocity  of  the  wind. 

83.  A  ship's  apparent  course  is  N.  33°  45'  E.,  8  knots  an 
hour ;  but  the  tide  sets  her  S.  67°  30'  E.  at  the  rate  of  3  knots 
an  hour.     What  is  her  true  course  and  rate  of  progress  ? 

84.  Two  forces,  P=  6.2  lbs.  and  Q  =  10.9  lbs.,  making  an 
angle  of  112°  4',  act  at  the  same  point  on  a  body.  Find  the 
magnitude  of  the  resultant  and  the  angle  which  it  makes 
with  the  force  P. 

85.  Three  forces  in  a  plane  acting  on  a  body  produce 
equilibrium.  The  greatest  is  5  lbs.,  the  smallest  3  lbs. 
The  angle  between  the  3  lb.  force  and  the  unknown  force 
is  79°  43'.     Find  the  third  force. 

86.  On  the  same  meridian  are  two  places  whose  difference 
of  latitude  is  68°  13'.  From  these  places,  the  zenith  dis- 
tances of  the  moon  at  its  culmination  were  found  to  be 
25°  and  44°  18',  respectively.  The  radius  of  the  earth  being 
3960  miles,  find  the  distance  of  the  moon  from  the  earth's 
center. 

87.  In  measuring  a  distance  AD,  a  portion  of  it  (BC) 
between  A  and  D  is  inaccessible.  AB  =  244,  CD  =  520. 
From  a  point  E  outside  of  AD,  angles  were  measured : 
AEB  =  82°  8',  BEC  =  21°  2',  CED  =  40°  14'.     Find  BC. 

88.  A  man  walking  along  a  straight  road  observes  the 
angle  of  elevation  of  a  tower  to  be  28°  40'.  At  a  point 
57  feet  further  on,  the  angle  of  elevation  of  the  same  tower 
is  48°  42'.  And  143  feet  beyond  that,  the  angle  is  33°  10'. 
Find  the  height  of  the  tower. 


90  OBLIQUE   TRIANGLES. 

89.  From  a  ship,  a  rock  is  seen  in  the  direction  N.  11°  48'  E. 
The  ship  sails  due  southeast  5  miles  when  the  direction  of 
the  rock  is  N.  1°  36'  E.  Find  the  distance  from  the  rock  to 
each  position  of  the  ship. 

90.  The  sides  of  a  triangle  are  three  consecutive  whole 
numbers;  and  the  greatest  angle  is  double  the  least  (A). 
Find  the  average  side  (c). 

91.  The  distance  between  the  centers  of  two  circles  is 
238  (c).  The  outer  common  tangents  make  with  each  other 
an  angle  of  36°  8'  (A).  The  inner  common  tangents  make 
with  each  other  an  angle  of  104°  12'  (B).  Find  the  length 
of  the  radius  of  the  larger  circle. 

92.  The  distances  between  three  points  A,  B,  C  are  as 
follows:  AC  =1200  feet,  BC=  1320  feet,  and  ^jB  =  1400 
feet.  An  observer  at  X  in  the  plane  ABC  finds  angle 
^X(7=:25°32'and  BXC  =  AO°.  Find  the  distances  of  X 
from  the  points  A,  B,  and  C,  if  C  and  X  are  on  different 
sides  of  AB  and  angle  AXB  =  AXC+  CXB. 

93.  From  the  top  of  a  cliff  300  feet  high,  the  altitude  of 
the  sun  was  observed  to  be  70°,  the  angle  of  elevation  of  a 
balloon  was  52°  12',  the  angle  of  depression  of  its  shadow 
on  the  sea  was  68°  14'  The  sun  is  in  front  of  the  observer, 
and  the  cliff,  sun,  balloon,  and  shadow  are  all  in  the  same 
vertical  plane.  Find  the  height  of  the  balloon  above  the 
sea. 

94.  From  a  bluff  100  feet  above  the  surface  of  a  lake,  the 
angle  of  elevation  of  a  balloon  is  48°  20',  and  the  angle  of 
depression  of  its  image  reflected  from  the  surface  of  the 
water  is  39°  50'.  Find  the  height  of  the  balloon  above  the 
lake. 

95.  From  the  deck  of  a  ship  sailing  due  east,  two  con- 
spicuous headlands  are  observed  to  bear  63°  and  24°, 
respectively,  to  the  northward  of  the  ship's  course.     After 


APPLICATIONS.  91 

sailing  8  miles,  the  corresponding  angles  were  found  to  be 
150°  and  98°.  Find  the  distance  of  the  headlands  from  one 
another. 

96.  Calculate  the  distance  of  two  inaccessible  points  A 
and  B,  knowing  a  base  CD  =  600  feet,  the  angle  BCD  =  40°, 
ACD  =  69°,  ADC  =  38°  30',  and  the  angle  BDC  =  70°  30'. 

97.  Three  points  A,  C,  and  B  being  given  on  a  chart 
of  a  coast,  it  is  desired  to  determine  the  position  of  a 
fourth  point  D.  ^(7=1000  feet,  5(7=850  feet,  angle 
^C5=114°40',  angle  ADC=W\1\  angle  C/)i?  =  30°  9'. 
Find  CD.  [CD  is  between  AD  and  BD;  C*  and  D  are  on 
opposite  sides  of  AB.^ 

98.  From  the  top  of  a  lighthouse  103.5  feet  above  the 
sea,  the  horizontal  angle  between  two  boats  was  found  to 
be  41°  20',  and  their  angles  of  depression  were  13°  23'  and 
17°  41'  respectively.     Find  the  distance  between  the  boats. 

99.  A  balloon  is  observed  from  two  stations  1  mile  apart. 
At   the   first   station,  the 

horizontal    angle    of    the  ^^^ 

balloon    and    the     other  ^.'' 

station  =  82°,     and     the  ^-''        / 

elevation   of   the    balloon  ^-^^  / 

=  12°.       At    the    second  ^^-''  / 

station,  the  horizontal  an-         ^-''  / 

gle  between  the  first  sta- AV / ;^B 

tion  and  the  balloon  =  65°.        ^^^ 


C 


Find   the   height    of    the         .  ^\^  ' 

balloon.  \        /  y^ 

100.   To   determine   the  ^^J^^ 

height  BC  of  'd  tower,  two 
points  A  and  D  90  feet 

apart  were  taken  in  the  horizontal  plane  through  its  base. 
The  angle  BAD  =  56°  33',  BAC  =  65''  23',  BDC  =56°  26'. 
Find  BC. 


92  OBLIQUE  TRIANGLES. 

101.  To  measure  the  breadth  of  a  river  AD,  a  point  B 
was  taken  in  AD  prolonged.  From  B  a  base  line  BC  =  400 
feet  was  laid  off,  making  an  angle  of  95°  16'  with  AB.  At 
C,  the  angles  BCA  =  52°  48'  and  BCD  =  24°  39'  were  meas- 
ured.    Find  AD. 

102.  From  three  points  A,  D,  B  in  Si  horizontal  straight 
line,  AD  =  500  feet,  DB  =  300  feet,  the  angles  of  elevation 
of  a  tower  were  measured,  and  found  to  be  9°  30',  51°  14', 
and  27°  51 ',  respectively.     Find  the  height  of  the  tower. 

103.  Three  points  A,  D,  B  lie  in  a  straight  line.  AD =232, 
DB  =  80.  To  locate  a  fourth  point  C  in  the  horizontal 
plane  of  these  three,  the  angles  ACD  =  85°  12'  and  DCB  = 
46°  13'  were  measured.     Find  CA,  CD,  CB. 

104.  The  pedestal  of  a  monument  is  9  feet  high,  the 
column  is  8  feet  high,  and  the  statue  is  10  feet  high.  A 
person  is  in  such  a  position  that  he  sees  the  three  parts 
subtending  the  same  angle.  AVhat  angle  with  the  monu- 
ment is  made  by  the  line  joining  his  eye  and  the  base  of  the 
monument ;  and  what  is  the  height  of  his  eye  above  the  base  ? 

105.  Within  an  angle  A  =  73°  44',  a  point  is  chosen  whose 
distances  from  the  sides  of  the  angle  are  6  inches  and  8 
inches.  A  line  is  drawn  through  this  point  so  that  the 
portion  included  between  the  sides  of  the  angle  is  bisected 
at  the  point.     Find  the  area  of  the  triangle  thus  formed. 

106.  The  two  non-parallel  sides  of  a  trapezoid  are  7.5 
feet  and  6.3  feet.  The  angle  formed  by  their  continuations 
=  41°  21'.  The  base  of  the  trapezoid  is  10  feet.  Find  its 
area. 

107.  From  two  vertices  of  an  equilateral  triangle,  whose 
side  is  234  feet,  two  bodies  begin  to  move  simultaneously 
toward  the  third  vertex ;  one  with  a  velocity  of  4  feet  per 
second,  the  other  with  a  velocity  of  3  feet  per  second.  When 
will  the  distance  between  the  moving  points  be  equal  to  the 
altitude  of  the  triangle  ? 


APPLICATIONS. 


93 


Fki.  5r 


108.  Calculate  the  radius  of  an  inaccessible  cylindrical 
tower  from  the  following 

data :  A  base  line  AB  is 
measured  =  70  feet.  The 
angles  formed  with  the 
base  line  by  the  pair  of 
tangents  are :  From  A, 
60°  and  20°;  and  from 
B,  75°  and  25°. 

109.  At  some  distance 
from  a  circular  pond,  a 
base  line  AB  =  250  feet 
is  measured.  At  A  the 
angles  between  the  base 

line  and  the  tangents  to  the  pond  are  71°  12'  and  54°  12'. 
At  B,  the  angle  between  the  base  line  and  the  tangent  to 
the  right-hand  side  of  the  pond  is  57°.  Find  the  radius 
of  the  pond. 

110.  At  some  distance  from  a  circular  pond,  a  base  line 
AB  =  75  yards  is  measured.  At  A  the  angles  made  by  the 
base  line  and  the  tangents  to  the  pond  are  72°  58'  and  44°  2'. 
At  B  the  angle  between  the  base  line  and  the  tangent  to 
the  left  side  of  the  pond  is  38°  20'.  Find  the  area  of  the 
pond. 

111.  To  find  tbe  height  of  a  hill,  three  points,  A,  J5,  and  C, 
not  in  a  straight  line,  were  found,  such  that  at  each  of  the 
three  points  the  angle  of  elevation  of  the  summit  of  the  hill 
was  the  same ;  viz.  52°  40'.  CB  was  measured,  and  found 
to  be  121.6  feet,  and  angle  BAC  =  15°  25'.  Find  the  height 
of  the  hill. 


CHAPTER   VI. 

PROPERTIES   OF  TRIANGLES. 


43.  The  altitudes.  — p  —  a  sm  B  =  h  sin  A.  Similar  ex- 
pressions hold  for  the  other  altitudes. 

If  the  three  altitudes  are  known,  the  triangle  can  be 
solved.     Denote  the  altitudes  by  pi,  p^,  and  p^.     Then 

Substitute  in  (21) ; 

P-2P3 

with  similar  expressions  for  cos  B  and  cos  C.     Then 
b  =  2^3  CSC  A;  c=p2  esc  A,  etc. 

The  bisectors  of  the  angles.  —  Let  CQ  (=  q)  be  the  bisector 
of  the  angle  ACB.     Then 

area  BCQ  +  area  QCA  =  area  BCA ; 

J  aq  sin  ^  C  +  ^  6g  sin  i  (7  =  ^  ah  sin  C  =  ab  sin  ^Ccos^  O. 

94 


ALTITUDES,   BISECTORS,  AND   MEDIANS.  95 

Whence  2a&cos-^C 

Now  angle  FOQ  =  \  {A—  B)-,  hence  q  =  i^  sec  -J-  {A  —  B). 
Hence  q  =  a  sin  B  sec  -|  {A  —  B). 

Similar  values  can  be  obtained  for  the  other  bisectors. 
AQ  =  b  CSC  ylQCsin  i  0;  QB  =  a  esc  CQB  sin  J  (7. 

Since  esc  ^QC  =  esc  CQB,    ^  =  -. 

QjB     a 

But    J(^4-Q^=c;  hence  ^(3  =  -^  and  QB        "^ 


rt  +  ^  a  -f-  6 

The  medians. — Let  M  be  the  middle  point  of  AB.    By  (20), 

a^  =  :^c^-\-  w?  —  cm  cos  BMC; 

6^  =  J  c^  +  wi^  —  cm  cos  ^  J/O. 
Now  cos  BMC  +  cos  ^4Af(7  =  0.     Add  and  solve  for  m. 


m  =  i^2a'  +  2b'-c\ 
Similar  relations  hold  for  the  other  medians. 
If  the  angles  are  known  and  the  medians  are  desired,  find 
the  angle  at  M  first.     Denote  the  middle  point  of  BC  by 
M„  the  middle  point  of  CA  by  M^.     Then  by  (24), 

cot  AMC  =  ^^^^  _  cot  A 

A  0 
=  l(GotB-cotA). 
Similarly,       cot  BM^A  =  J-  (cot  C  —  cot  B). 
cot  CM,B  =  I  (cot  A  -  cot  C). 
Check :  cot  AMC  +  cot  B3IiA  +  cot  CM,B  =  0. 

If  the  three  medians  are  known,  m^,  ?%,  m^,  the  sides  can 
be  found.  Their  values  can  be  obtained  from  these  equa- 
tions : 

9a'  =  Sm,.,'-\-8ms'-4:7n^% 

9b-r=8  7tii  +  8  m^'  -  4  m/, 
9c^==8;».,--h8m,--4m32. 


96 


PROPERTIES   OF  TRIANGLES. 


Fig.  57. 


44.  The  incircle  and  the  excircles.  —  Let  0  be  the  center 
of  the  incircle  touching  the  sides  of  the  triangle  ABC  in  Aq, 
Bq,  and  Co ;  Oj,  the  center  of  the  excircle  in  angle  A  and  A^, 
Bi,  Ci,  its  points  of  tangency,  etc.     Then 

AOi  =  AB,  =  BA,  =  BC2  =  CBs  =  CA^  =  s, 
ACo  =  ABo  =  BA^  =  BCs  =  CB^  =CA,  =  s-a, 
ACs  =  AB^  =  BAo  =  BCo  =  CB^  =CA,  =  s-b, 
AC2  =  AB2  =  BA^  =  BC^  =  CBo  =  OAo  =  s-c. 

Denote  the  radius  of  the  incircle  by  r,  the  radii  of  the 
excircles  by  Vi,  r^,  and  r^  respectively. 

Values  of  the  various  segments  in  the  figure  will  now  be 
obtained  in  terms  of  the  sides. 

OCo=  05 sin  i 5;   OB  =  a  sin  ^  C esc BOC(=  90° -h^ A), 
r  =  a  sec  J  A  sin  i  B  sin  i  C 
=  b  sin  ^A  sec  1 B  sin  ^  C 
=  c  sin  ^  J.  sin  1  i?  sec  i  C. 


THE   INCIRCLE   AND   EXCIHCLES.  97 

Since  OjOi  =  OiB  cos  |  B, 

and  OiB  =  acsc  GO,B{=  90°  -\A)  sin  0,CB{=  90°  -  ^C). 
Also,  OiiB  =  c  CSC  ^Oi5(=  1  C)  sin  1 A 

ri  =  a  sec  ^  ^  cos  |  B  cos  ^  C 
=  6  sin  J  ^  esc  J-  J5  cos  \  C 
=  c  sin  ^  ^  cos  I JB  esc  J-  C. 

Find  02C2(=  r^  and  03(73(=  ?3)  in  a  similar  manner. 

Since  ACi  =  Vi  cot  \  A. 

s  =  tt  esc  I  ^  cos  \  Bcos^C 
=  b  cos  ^A  CSC  ^B  cos  ^  C 

=  C  COS  1^  ^  COS  I  5  CSC  ^  C. 

ACo  =  rcot\A;  BCo  =  r  cot  ^B-,  CA)=r  cot  ^C. 

s  —  a  =  a  csc^Asm^  Bsin^C 
=  b  cos  ^  ^  sec  J  J3  sin  ^  C 
=  c  cos  ^Asin^B  sec  |  C. 

s  —  b  =  asec^A  cos ^ ^ sin ^  C 
=  6  sin  ^  ^  esc  1^  J5  sin  i  (7 
=  c  sin  1^  ^  cos  ^Bsec^  C. 

s  —  c=aseG^Asm^Bcos^C 
=  6  sin  I  ^  sec  I  ^  cos  \  C 

—  c  sin  ^Asin^B  esc  ^  C. 
^0=(s-a)seci^;l?0==(«-?>)sec|5;  00=  (s-c) sec  1  C. 
^Oi=ssec^^;  ^'Oi=(.s-c)csc  JJS;  OOi =(«-?/)  esc  }  O. 
^02=(s-c)csci^;  jB02=sseci7^;  00,= (s- a)  esc  1  0. 
^03=(s-&)csc^^;  ^03=  (6*— a)  esc  IB;  003=sseeiO 
00i=^0i-^0=aseei^;  OO.^ft  sec iB;  003=cseciO. 
Oa03=^02+^03=a.csc|-^;  030i=6csc|-5;  Oi02=ccsciO. 
Angle 
O3OiO2=90°-i^;  OiO2O3=90°-i5;  0^0301= 90° -iO 


propp:rties  of  triangles. 


Functions  of  the  half-angles  of  a  triangle. 

'  be  '  2  (^s-b)(s-c)' 

s(.s-a)  ^  be 

AVith  similar  relations  for  i  jB  and  4  (7. 


45.  The  circumcircle.  —  Let 
F  be  the  center  of  the  cir- 
cumcircle, and  M  the  middle 
point  of  AB.  Then  since 
angle  AFB  at  the  center  of 
the  circle  is  twice  the  angle 
C  on  the  circumference, 
angle  AFM=  C. 

Now  AF=AMcsgAFM. 
Therefore,  changing  letters, 

i2  =  i  a  CSC  ^  =  i  6  CSC  5 


Now 

Also, 
the  sum  and  product  of  which  give 

7^  (sin  A  +  sin  B  -|-  sin  C)=  s, 
Tl^  sin  A  sin  i?  sin  (7  ==  J  ahc  =  i  RK. 

46.  Relations  between  the  sum  or  difference  of  two  sides  and 
the  sum  or  difference  of  two  angles. 

Let  ABC  be  the  triangle  under  consideration.  With  C  as 
a  center  and  radius  equal  to  CA  (b),  describe  a  semicircle, 
cutting  BC  in  JJ  and  E.     Then  BD  =  a-b  and  BE  =  a-\-b. 


i  c  CSC  C. 

Fig 

.  58. 

K  = 

i  be  sin 

A  = 

rs. 

Hence 

R  = 

abc 

abe 
4  rs 

R 

sin^  = 

ia, 

R 

sin 

B  =  i?>, 

R  sin  C  = 

ic; 

DIFFERENCE   OF   TWO  ANGLES.  99 


Angle  DAE  =  90°  (in  semicircle),  angle  E  (on  circnmfer- 
ence)=  |  C,  angle  BDA  (exterior)  =  90°  -f  |  0,  angle  DAB 
=  180°  ~B- (90°  +  ^C)=^(A-B). 
In  triangle  ^i5/>,  by  (19),  AB  sin  Z>^5  =  DB  sin  ^Z)2?. 

c  sin i(^  -B)  =  (a-b)  cos  ^  (7.  (31) 

In  triangle  AEB,  by  (19),  AB  sin  ^^5  =^B  sin  E. 

c  cos  1  (vl  -  JB)  =  (a  -h  b)  sin  |  O.  (32) 

The  quotient  of  (31)  and  (32)  reduces  to 

(a  +  b)  tani(yl  -  B)  =  (a  -  &)tan  i(^  +  B). 
The  product  and  difference  of  the  squares  reduce  to 
c^sin  {A  -  B)  =  (a'  -  b")  sin  G. 
c^cos  {A  -  B)=2  ab  -  (a"  -f-  b')  cos  0. 
Eliminate  a^  by  (20),  and  a  and  b  by  (19). 

sin  (^  -  JS)  =  sin  C  -  2  sin  B  cos  A,  (33) 

cos  (^  —  7?)  =  2  sin  ^  sin  jB  -  cos  C.  (34) 

The  other  functions  of  ^  —  J5  can  then  be  found.     Since 

A-\-  B  =  180°  —  C,  any  function  of  A -\-  B  can  be  expressed 

in  terms  of  functions  of  C. 

Also,  since   B  +  ^C  =  90°  -  ^(A  -  B),  any  function  of 
B  -\-  ^C  is  equal  to  the  cofunction  of  ^(A  —  B). 


100  PROPERTIES  OF  TRIANGLES. 

47.  General  method  of  solving  a  triangle,  given  three  seg- 
ments in  or  about  the  triangle.  —  Suppose  that  to  solve  the 
triangle  there  is  given  any  three  segments  (M,  r,  r^,  s,  s  —  a, 
etc.).  Each  of  these  segments  may  be  expressed  in  terms 
of  the  sides  and  functions  of  the  half  angles ;  these  func- 
tions may  be  expressed  in  terms  of  the  sides.  Hence  the 
sides  may  be  found  by  solving  the  set  of  equations  thus 
obtained. 

Relations  between  the  segments.  —  By  using  values  of  any 
two  of  these  segments  a  relation  between  the  segments  may 
be  obtained.     For  example,  find  r  and  s  in  terms  of  M. 
r  =  a  sec  i  J.  sin  i  B  sin  a  C. 
a  =  2  jR  sin  J.  =  4  i?  sin  i  ^  cos  ^  A. 
Hence         r  =  4  i2  sin  i  J.  sin  i  JB  sin  i  C. 
Similarly,   s  =  iB  cos  i  A  cos  ^  Bcos^  C. 

48.  Points  of  intersection  of  the  circumcircle.  —  Let  tri- 
angle ABC,  Fig.  GO,  be  the  triangle  under  consideration,  the 
bisectors  of  the  interior  and  exterior  angles  being  drawn  as 
in  Fig.  57. 

Also,  let  F  be  the  center  of  the  circumcircle  of  the  tri- 
angle ABC.  Since  angle  M"'CO  is  a  right  angle,  it  is  in- 
scribed in  a  semicircle ;  hence  M"'F  passes  through  K'", 
the  intersection  of  the  circumcircle  and  CO3.  The  lengths 
of  OoM'"  and  of  O^K'"  will  now  be  determined. 
Since  FCB  =  90°  -  ^  (see  Fig.  58), 

angle  FCK'^'  =  CK"'F=  lC-(90°  -  A)=^{A  -  B). 
CM'"  =2Rsm^(A-  B),  CK"'  =  2  i2  cos  i(^  -  B). 
But  i?  =  i  c  sec  i-  O  CSC  i  C, 

c  sin  ^(A  —  B)  =  (a  —  b)  cos  J  C, 
c cos  i(^  -B)  =  {a  +  b)  sin  |  C. 
CM'"  =^(a-b)  CSC i  C;     CK'"  =  i(a  -f  6)  sec  1  C. 
O2M'"  =:  i  c  CSC  i  (7 ;  OsK"'  =  i  c  sec  1  (7. 


INTERSECTIONS  OF  THE   CIRCUMCIRCLE. 


101 


Whence  M"  and  A'"'  are  the  middle  points  of  Ofi,  and  00 ^^ 

respectively. 

In  the  same  manner  it  may  be  shown  that  M',  K',  M",  K", 
are  the  middle  points  of  O2O3,  OOj,  0^0^,  OO2,  respectively. 


Fig.  60. 


If  OCq(=  r)  is  perpendicular  to  AB,  the  angle  CqOK'"  is 
also  equal  to  ^{A  —  B),  and,  therefore,  K"'F  is  parallel  to 
OCq',  it  is,  therefore,  perpendicular  to  AB  through  its 
middle  point  M.     Hence,  K'"  is  the  middle  point  of  the 


102 


PROPERTIES   OF  TRIANGLES. 


arc  AB,  K'  is  the  middle  point  of  the  arc  BC,  and  K"  is 
the  middle  point  of  the  arc  CA. 

The  circiimcircle  of  ABC  is  called  the  nine-points  circle  of 
the  triangle  O1O2O3. 


49.  The  nine-points  circle.  —  Let  ABC  be  the  triangle 
under  consideration.  Draw  the  altitudes  AP^,  BP^,  CPs, 
meeting  in  H.     Join  P1P2,  AA>  Ps^i- 

In  the  triangle  AP^C,      AP^  =  b  cos  A, 
and  in  triangle  AP2B,          AP^  =  c  cos  A. 

Therefore  the  triangles  APiP^  and  ABC  having  a  com- 
mon angle  and  the  including  sides  in  each  triangle  propor- 
tional are  similar ;  hence    P2P3  =  «  cos  A. 

In  a  similar  manner   P^Px  =  h  cos  B  and  P^P^  =  c  cos  C 


THE  NINE-POINTS  CIRCLE.  103 

Also  .  angle  B1\P^  =  CP.Po  =  A,  whence  P,PJ\  = 
180°  -2  A,   P.P^P;,  =  180°  -2B,   P.P^P,  =  180°  -  L>  C, 

The  intersection  //  of  the  altitudes  is  called  tlie  ortho- 
center  of  the  triangle  ABC\  the  triangle  PJ\P.^,  joining 
the  feet  of  the  altitudes,  is  called  the  pedal  triangle  of  the 
triangle  ABC.  It  should  be  noticed  that  the  orthocenter 
of  a  triangle  is  the  center  of  the  incircle  of  the  pedal 
triangle.  The  circumcircle  of  the  pedal  triangle  is  the  nine- 
points  circle  of  the  original  triangle. 

All  of  the  properties  of  any  triangle  ABC  can  be  reduced 
to  properties  of  its  pedal  triangle  P1P2P3  by  substituting 
in  any  formula  referring  to  the  triangle  ABC,  180°  — 2^1 
wherever  A  occurs,  180°  —  2B  wherever  B  occurs,  180°  — 
2  C  wherever  C  occurs.  After  completing  these  substitu- 
tions, put  a  cos  A  wherever  a  occurs,  b  cos  B  wherever  b 
occurs,  and  c  cos  C  wherever  c  occurs.  For  instance,  make 
these  substitutions  in  the  value  of  U.  The  radius  of  the 
circumcircle  of  PiPJ\  or  the  nine-points  circle  of  ABC 
=  i  a  cos  ^  sec  i  (180°  -2  A)  esc  i  (180°  -  2  .1) 
=  ^  a  CSC  ^  =  ^  7?. 

That  is,  the  radius  of  the  nine-points  circle  of  any  triangle 
is  one-half  the  radius  of  the  circumcircle  of  the  triangle. 

All  other  properties  of  the  nine-points  circle  of  OiOM^ 
(circumcircle  of  ABC)  can  be  reduced  to  properties  of  the 
nine-points  circle  of  ABC  (circumcircle  of  P^P^P:^  by  the 
same  substitutions. 

If  F  is  the  center  of  the  circumcircle  (see  Fig.  58), 

K  =  area  BFC  +  area  CFA  +  area  AFB, 

that  is,      rs  =  ^ aR  cos  A  -{■  \bR  qo&  B  -\-  \ cR  cos  C. 

„  2  s  2irR 

Hence         = 

a  cos  A-\-b  cos  B -{-  c  cos  C      2  irr 

That  is,  the  perimeter  of  the  triangle  is  to  the  perimeter 
of  the  pedal  triangle  as  the  circumference  of  the  circum- 
circle is  to  the  circumference  of  the  incircle. 


104  PROPERTIES  OF  TRIANGLES. 

50.  Abscissae  and  ordinates  of  various  centers.  —  Count 
abscissae  from  A  toward  B. 

For  F,  the  center  of  the  drcumcircle. 

The  abscissa  is  ^  c  and  the  ordinate  is  ^  c  cot  G. 

For  0,  the  center  of  the  incircle. 

The  abscissa  is  s  —  a  and  the  ordinate  is  (s  —  a)  tan  i  A 

For  Oi,  the  center  of  the  excircle  in  A. 

The  abscissa  is  s  and  the  ordinate  is  s  tan  i  A. 

For  O2,  the  center  of  the  excircle  in  B. 

The  abscissa  is  —  (s— c)  and  the  ordinate  is  (s— c)cot^^. 

For  O3,  the  center  of  the  excircle  in  C. 

The  abscissa  is  s—b  and  the  ordinate  is  — (s— 6)cot^A 

For  //,  the  iyitersection  of  the  altitudes  (prthocenter). 

The  abscissa  is  h  cos  A  and  the  ordinate  is  h  cos  A  cot  B. 

For  6r,  the  intersection  of  the  medians. 

The  abscissa  is  \{h  cos  A+c)  and  the  ordinate  is  ^  &  sin  A. 

For  N,  the  center  of  the  nine-points  circle. 

The  abscissa  is  ^  6  cos  A-\-\c  and  the  ordinate  is  |^  5  sin  A 
—  \c  cot  (7. 

To  obtain  the  abscissa  and  ordinate  of  N,  note  that  the 
angle  BP^N  is  90°  —  (^  —  B).  Then  these  values  can  be 
deduced  from  (33)  and  (34). 

To  obtain  the  distance  between  any  two  of  these  points,  sub- 
tract their  abscissa  and  square  the  difference ;  subtract  their 
ordinates  and  square  the 
difference.  The  square 
root  of  the  sum  of  the 
squares  gives  the  required 
distance. 

These  distances  can  also 
be  obtained  from  the  tri- 
angle. Thus  to  find  the 
distance  FH,  find  AH,  AF,  angle  FAH,  and  apply  (20) : 

AH  =h(iosA  sec  (90°  -  B)  (See  Fig.  61) 

=  h  CSC  B  cos  A=  a  esc  A  cos  A=  a  cot  A. 


Fia.  62. 


DISTANCES  BETWEEN  CENTERS.  106 

Since  a  =  2  Esin  A,     AH  =  2  R  cos  A. 
Angle       BAH  =  90°  -  jB ;  BAF  =  90°  -  C; 
hence  angle  FAH=  C  —  B. 

Hence  FH^  =  i?^  [1  +  4  cos^  ^  -  4  cos  (0  -  B)  cos  A']. 
In  (34)  put  90°  -  C  for  A,  90°  -  B  for  ^,  and  180°  -  A 

cos  {C—  B)  =  2  cos  O cos  5  +  cos  A. 
Hence       i^//^  =  R^{1-S  cos  ^  cos  5  cos  C). 

To  find  (?F  and  (?/. 

vlO=r CSC  L^  and  angle  F^O=J^-(90-C')  =  ^(C-fi). 

OF'=R-  +  r^csc^l^  -  2  72r  csc  ^^  cos  ^  (0  -  i^). 

Since  r  =  4  72  sin  |  ^  sin  ^  ^  sin  ^  C, 

r" csc2 1  ^  =  4  i^r  CSC  ^ .4  sin  ^  jB  sin  1  (7  =  2  i?r ^^^~^X 

a 

6  +  c 


By  (32),         CSC  ^  ^  cos  ^  (C  -  ^)  = 


a 


Substituting  OF'-  =  R^-2Rr. 

Similarly,  O^F^  =  R^  +  2  Rr^. 

Exercises:  —  Show  that  OH  =2i'^  —AR^cosA  cos  B  cos  C. 
ON=iR-r. 

0,N=iR-^n. 

PROBLEMS. 

51.  112.  The  radius  of  the  incircle  of  a  triangle  is  6; 
^  =  38°  53',  0  =  73°  44'.  A  line  is  drawn  across  angle  A 
parallel  to  the  side  BC  and  tangent  to  the  incircle.  In  the 
small  triangle  thus  formed,  a  circle  is  inscribed.  Find  the 
radius  of  this  latter  circle. 

113.  In  the  triangle  of  the  last  problem  (r=6,  ^=67°  23% 
a  small  circle  is  drawn  touching  each  side  of  angle  A,  and 


106 


PROPERTIES   OF   TRIANGLES. 


also  tangent  externally  to  the  incircle.     Find  the  radius  of 
this  small  circle. 

114.   In  the  same  triangle  (r  =  6,A  =  67°  23',  B  =  38°  53'), 

the  incircle  touches  the  sides  at  A^,  li^,  and  Cq.     Find  the 
radius  of  the  incircle  of  the  triangle  A^,  B^,  C\. 

Malfatti's  problem.  —  To  inscribe  in  a  triangle  three  circles, 
each  touching  the  other  two  and  also  two  sides  of  the  tri- 
angle.* 

Let  the  sides  of  the  triangle 
be  a  =148,  b  =  175,  c  =  153, 
and  the  angles  A  =  53°  8', 
i^=71°4',  0=55°  48'. 

Denote  the  radii  of  the  Mal- 
fatti   circles   by  x,   y,  and   z. 

Hence 
ic  cot  \A  -\-  2  -\/xy  4-  y  cot  \B—c. 

Multiply  by   sin  i  B  cos  }jB  =  \  sin  B. 
X  cot^  A  sm^  B  CQs^  B  +  2  sink  B  cosh  B  ^  xy  -{-yzos^  -2  B =^csinB . 


*  Steiner  has  given  the  following 
construction  for  tliis  problem.  Let 
0  be  the  center  of  the  incircle  of 
ABC.  Through  the  point  of  con- 
tact D  of  the  incircle  of  AOB,  draw 
a  tangent  to  the  incircle  of  BOC, 
meeting  BC  m  F.  The  incircle  of 
the  triangle  BDF  is  a  Malfatti  cir- 
cle. Each  of  the  other  two  Malfatti 
circles  is  constructed  in  a  similar 
manner. 


MALFATTI'S   PROBLEM.  107 

o  ri 

The  coefficient  of  x  equals  sin^  ^B-\ ; 

a 

also  c  sin  B  =  pi.     Hence 

( V^  sin  ^  J3  +  Vy  cos  i  By  =  ^}h-  ^''  ~  ''^^- 

When  y  is  changed  to  2;,  B  is  changed  to  C 

( V^  sin  i  O -f- V^  cos  i  Cf  =  ^i),  -  ^-^ 

Equating  and  changing  letters, 

^/x  sin ^B  +  Vy  cos -^  J5  =  Vx  sin |^  6^  +  V^  cos ^  C. 

V//  sin  ^  O  +  V^  cos  ^  C  =  Vy  sin  ^  .4  +  Va;  cos  ^  ^. 

Adding  these  and  arranging, 

V.V _ cos ^A  —  sin^  B  +  sin^  C _  ^..^ 
-y/^ ~ cos ^B  —  sin ^ yl  -h  sin  1^  C 

Substituting,    a;  =  30,    2/ =  26.28,    2  =  28.01. 

Properties  of  the  Malfatti  radii.  —  In  the  equation* 

sin  A  +  sin  i^  +  sin  C  =  4  cos  ^Acos^B  cos  ^  C, 

put  90°  -  i  ^  instead  of  A,  —  ^  B  instead  of  B,  and  180° 
—  ^  O  instead  of  C;  then  change  A  and  i?: 

cos  4^  —  sin  ^  5  4-  sin  ^  C  =  4  cos  (45°  —  :J^  ^)  cos  :^  2^  sin  ^  C, 

cos  ^  ^  —  sin  ^  ^  4-  sin  ^  C  =  4  cos  J  ^  cos  (45°  —  -J-  B)  sin  \  C. 

Hence  ^  =  ^""^  (—  ~^^^  ^Q^?-^ 

V^     cos  \  A  cos  (45°  —  \B) 

Now  cos  (45°  —  J  ^)  =  cos  45°  (cos  \A-[-  sin  \  A),-\ 

Also  cos  (45°  -\B)  =  cos  45°  (cos  i  i5  -f  sin  1  B). 

Hence 

V^(l  +  tani^)  =  V^(l  +  tan}5)  =  V2(l  +  tani  C). 

*  Proved  on  next  page. 

t  See  equation  (68),  Appendix  II,  page  145. 


108  PROPEKTIES  OF  TRIANGLES. 

52.  Relations  between  the  angles  of  a  triangle.  —  When  two 
relations  connecting  the  same  two  segments  are  obtained, 
their  comparison  gives  a  relation  between  the  angles  of  the 
triangle. 

For  example,  72  (sin  ^  -f-  sin  JB  +  sin  C)=  s. 

s  =  4  E  cos  1 A  cos  ^  B  cos  i  C. 
sin  ^  +  sin  i5  +  sin  C  =  4  cos  ^^  A  cos  }^  B  cos  \  C. 

And  since  the  only  condition  on  A,  B,  and  C  is  that  their 
sum  is  180°,  this  equation  is  true  for  any  three  angles  having 
this  sum.     Thus,  apply  it  to  the  pedal  triangle, 

180°  -  2  ^  for  A,    180°  -  2  J3  for  B,    180°  -  2  C  for  C. 
sin  2  A-\-  sin  2  J3  +  sin  2  C  =  4  sin  A  sin  B  sin  C. 

By  adding  and  multiplying  equations  in  (22), 

coti^  +  cot|J?  +  cotiC  =  ^^-^"  +  ''  +  ''>  =  g; 

TV 

cot  i  ^  cot  i  B  cot  1  0  =  (i^l^5fc*l(^^^  =  ^  =  ?. 

Hence 

cot  i  ^  +  cot  1 JB  -|-  cot  ^  C  =  cot  i  ^  cot  ^  B  cot  1  C. 
In  this,  put 
180°  -2  A  for  A,   180°  -2B  for  B,   180°  -  2  C  for  C. 

tan  A  +  tan  B  +  tan  O  =  tan  A  tan  _B  tan  C. 
Divide  this  equation  by  the  right-hand  member, 
cot  B  cot  (7  +  cot  O  cot  J.  4-  cot  ^  cot  J5  =  1. 

It  will  be  seen  that  there  is  no  limit  to  the  number  of  re- 
lations which  can  thus  be  derived. 

Exercises.  — For  all  triangles,  prove 

sin^  A  -f  sin^  B  +  sin^  O  =  2  -f-  2  cos  A  cos  B  cos  C, 
sin^  A  -f  sin^  B  +  cos^  O  =  1  -f  2  sin  ^  sin  B  cos  C, 
cos  A  -f-  cos  B  -f  cos  C    =  1  +  4  sin  i  ^  sin  i  B  sin  i  (7, 
cos  2  J.  +  cos  2  ^  -f  cos  2  C  =  —  1  —  4  cos  y1  cos  5  cos  C. 


ELEMENTS  OF  OBLIQUE  TRIANGLES. 


10^ 


Exercises.  —  The  following  table  contains  elements  of  six 
different  triangles.  Assume  three  independent  elements  in 
any  one  of  the  triangles  and  compute  other  elements. 


a 

1 

2 

3 

4 

5 

6 

13 

25 

37 

37 

68 

68 

h 

15 

26 

13 

15 

(j^y 

75 

<• 

14 

17 

40 

44 

57 

77 

(^i 

9 

10 

5 

9 

25 

45 

€■2 

5 

rr 
i 

35 

35 

32 

32 

.S 

21 

34 

45 

48 

95 

110 

s  —  a 

8 

9 

8 

11 

27 

42 

s-b 

6 

8 

32 

33 

30 

35 

s—c 

7 

17 

5 

4 

38 

33 

A 

53°  8' 

67°  23' 

67°  23' 

53°  8' 

67°  23' 

53°  8' 

B 

67°  23' 

73°  44' 

18°  55' 

18° 55' 

61°  56' 

61°  56' 

G 

59°  29' 

38°  53' 

93°  42' 

107°  57' 

50° 41' 

64°  56' 

K 

84.00 

204.0 

240.0 

264.0 

1710 

2310 

R 

8.125 

13.54 

20.04 

23.12 

36.83 

42.50 

r 

4.000 

6.000 

5.333 

5.500 

18.00 

21.00 

n 

10.50 

22.67 

30.00 

24.00 

63.33 

55.00 

r2 

14.00 

25.50 

7.500 

8.000 

57.00 

66.00 

n 

12.00 

12.00 

48.00 

66.00 

45.00 

*70.00 

Pi 

12.92 

16.32 

12.97 

14.27 

50.29 

67.94 

Pi 

11.20 

15.69 

36.92 

35.20 

52.62 

61.60 

Ps 

12.00 

24.00 

12.00 

12.00 

60.00 

60.00 

Qi 

12.95 

17.11 

16.33 

20.01 

50.53 

67.96 

(J2 

11.22 

16.19 

37.92 

39.65 

53.18 

61.93 

Qz 

12.09 

24.04 

13.16 

12.56 

60.07 

60.18 

mi 

12.97 

22.49 

23.29 

27.17 

50.80 

68.09 

m2 

11.24 

20.33 

37.98 

39.95 

53.67 

64.02 

ms 

12.17 

24.19 

19.21 

17.69 

60.10 

61.39 

CHAPTER   Vir. 

SPHERICAL   TRIANGLES. 

53.  If  there  are  three  points  on  the  surface  of  a  sphere, 
and  each  pair  of  points  is  joined  by  an  arc  of  a  great  circle, 
the  surface  of  the  sphere  is  divided  into  two  parts,  each  of 
which  is  called  a  Spheri-  q 

cal  Triangle.     Join  each  ^--'''^^"Vx 

of  the  three  points  to  0,  ^^^^"""^  \  ^ 

the  center  of  the  sphere.  ^^^^-^"^^^  \ 

The    arcs     joining     the    C><^ 4b 

pairs   of   points   on   the  ^^"-^^^^^  /      / 

surface  measure  respec-  ^"^-^-..^^^  /  / 

tively  the  plane  angles  ^^^--.>^ 

formed  at  the  center  of 
the   sphere.      Thus    the 

arc  AB  measures  the  plane  angle  AOB.  Also  the  angles 
between  the  arcs  are  equal  respectively  to  the  dihedral 
angles  made  by  the  planes  of  the  arcs.  Thus  the  angle  A 
of  the  spherical  triangle  is  equal  to  the  angle  between  the 
planes  ACO  and  ABO. 

There  is  no  limit  to  the  number  of  different  possible 
values  of  any  side  or  angle  of  a  spherical  triangle.  For  the 
arc  joining  A  and  B  may  be  any  arc  AB  +  any  number  of 
circumferences.  And  the  angle  A  may  be  generated  by 
turning  AQ  away  from  AB  through  any  angle  A  +  any 
number  of  revolutions.  Since  any  function  of  n  x  360°  +  x 
is  equal  to  that  function  of  x^  it  follows  that  any  single 
solution  of  a  spherical  triangle  includes  the  solution  of  all 
spherical  triangles  which  have  any  side  or  angle  increased 

110 


THE   GENERAL  TRIANGLE. 


Ill 


by  any  number  of  times  360°.  Therefore,  without  loss  of 
generality,  each  side  and  each  angle  may  be  considered  less 
than  360°. 

Taking  each  side  less  than  360°,  there  may  be : 


Fi<>.  G.5. 
ao  =  a      Ao  =  300'^  —  A 
bo  =  b       Bo  =  360°  -  B 
co  =  c       Co  =  SGir  -  C 

1.    None  more  than  180°. 


Fig.  m. 
ai^nGO^-a    y1i  =  3r,0°-^ 
bi  =  b  Bi=  180°  -  B 

ci.=  c  C'l  =  180°  -  C 

2.    One  more  than  180°. 


Fig.  G7.* 
ao  =  3G0°  -a     Ao 
h.2  =  300°  -  h     Bo 

C2  =  C  Ci 

3.    Two  more  than  180°. 


Fig.  68.* 

180°- 

A 

as  =  300°  -a    As  =  A 

180°- 

B 

bs  =  300°  -b     Bz  =  B 

300°  - 

-  C 

C3  =  360°  -  c      t\=  C 

80°. 

4. 

Three  more  than  180°. 

*  Out  of  perspective.     It  will  be  well  if  all  the  diagrams  in  this 
chapter  are  drawn  with  colored  chalk  on  a  spherical  blackboard. 


112 


SPHERICAL  TR'TANGLES. 


The  four  triangles  in 
Eig.     69     are     called 
Associated     Triangles. 
The  three  triangles  as- 
sociated with  ABC  are 
obtained   by  prolonging 
the  sides  of  ABC. 

Spherical  triangles, 
like  plane  triangles,  are 
right,  isosceles,  equi- 
lateral, and  oblique ;  but 
a  right  spherical  triangle 
may  have  more  than  one 
right  angle.     If  one  or 


more  sides   of   the  triangle  is   a 


quadrant,  the  triangle  is  said  to  be  quadrantal. 


FORMULAE   FOR   RIGHT  TRIANGLES. 

54.  Let  ABC  be  a  spherical  right  triangle,  right  angled 
at  B.  Let  the  angles  A  and  C  be  acute,  and  each  side  of 
the  triangle  be  less  than 
180°.  Join  A,  B,  and  C 
to  0,  the  center  of  the 
sphere.  Then  the  planes 
AOB  and  BOC  are  at 
right  angles.  Erom  any 
point  D  in  the  edge  0(7, 
drop  a  perpendicular 
D^i  to  the  plane  AOB. 
Through  DB^  pass  a 
plane  at  right  angles  to  the  edge  OA.  Since  DB^  is  per- 
pendicular to  the  plane  AOB,  it  is  perpendicular  to  OBi 
and  BiAi',  and  since  AiBi  and  AiD  are  both  perpendicular 
to  OA,  the  angle  B^AJJ  measures  the  angle  between  the 
planes  AOB  and  AOC. 


Fig.  70. 


FORMULAE  FOR   RIGHT  TRIANGLES.  113 

OD  sin  a  =  B^D  =  A^D  sin  A ;    and  A^D  =  OD  sin  6. 
Therefore  sin  a  =  sin  6  sin  >l.  (35) 

OBi  tan  a  =  B^D  =  AiBi  tan  A ;  and  AiBi  =  OBi  sin  c. 
Therefore  tan  a  =  sin  c  tan  A.  (36) 

OAi  tan  b  =  AJ)  =  AiBi  sec  A ;  and  ^jSi  =  OAi  tan  c. 
Therefore  tan  6  =  tan  c  sec  /!•  (37) 

If  the  angle  A-^B^Cs,  Fig.  69,  is  a  right  angle,  these  for- 
mulae (35),  (36),  and  (37)  may  be  shown  to  apply  to  the  tri- 
angle A^BqCs.  Thus  put  for  a  the  value  180°  —  ttg,  put  for 
b  the  value  180° -63,  put  for  A  the  value  180°— ^3.  There 
is  obtained: 

sin  (180°  -  a,)  =  sin  (180°  -  63)  sin  (180°  -  ^3)  ; 
whence  -  sin  a^  =  sin  63  sin  A^. 

tan  (180°  -  as)  =  sin  c  tan  (180°  -  ^3)  ; 
whence  tan  a^  =  sin  C3  tan  A^. 

tan  (180°  -  63)  =  tan  c  sec  (180°  -  ^3)  ; 
whence  tan  b^  —  tan  C3  sec  A^. 

It  therefore  follows  that  the  formulae  (35),  (36),  (37)  are 
true  when  one  of  the  angles  of  the  right  spherical  triangle 
is  obtuse.  In  a  similar  manner,  these  formulae  may  be 
shown  to  apply  to  the  triangle  ^2-S2C2,  Fig.  69,  and  are 
therefore  true  when  two  of  the  angles  are  obtuse. 

It  therefore  follows  that  formulae  (35),  (36),  (37)  are  true 
for  all  right  triangles  in  which  each  of  the  parts  is  less 
than  180°. 

In  the  same  three  formulae  put  successively  the  values 
of  the  parts  of  the  various  triangles  in  Figs.  66  to  68,  in 
which  assume  B  =  90°.  It  will  be  found  that  all  three 
formulae  apply ;  they  are  therefore  true  for  all  right  spheri- 
cal triangles. 


114 


SPHERICAL  TRIANGLES. 


Collecting, 


Changing  letters, 


Sq.  root  product  (35)  to  (40), 
(41)  --  product  (36)  and  (39), 
(35)  X  (40)  -f-  (42)  gives 
(37)  X  (38)  ^  (42)  gives 


sin  a 
tana 
tan& 
sine 
tan  c 
tan  h 
sec  h 
sec  h 
sec  c 
sec  a 


sin  h  sin  A, 
sin  c  tan  A, 
tan  c  sec  ^. 

:  sin  h  sin  C, 
sin  rt  tan  C, 
tan  a  sec  (7. 

:  tan  A  tan  (7. 

:  sec  a  sec  c. 
sin  A  sec  C 

:  sec  A  sin  C. 


(35) 
(36) 
(37) 
(38) 
(39) 
(40) 
(41) 
(42) 
(43) 
(44) 


55.  The  following  diagrams  with  the  accompanying  rules 
will  be  found  useful  in  rapidly  picking  out  the  relation 
which  exists  between  any  three  parts  of  a  spherical  right 
triangle. 


sin  a — sin  A-     ^sec  G- 


a-<osQr 


FiG.  71. 

Rule  I.  Any  function 
is  equal  to  the  product  of 
its  adjacent  functions. 

Rule  II.  Any  function 
is  equal  to  the  product  of 
adjacent  function  and  the 
reciprocal  of  the  next  func- 
tion. 


mQ 


Fig.  73. 


SOLUTIONS.  115 

56.  In  the  discussion  of  plane  right  triangles  it  was  found 
that  any  two  parts,  provided  one  of  them  was  a  side,  was 
sufficient  to  make  the  solution  of  a  right  triangle  deter- 
minate. In  spherical  triangles  with  one  right  angle,  there 
will  always  be  two  solutions  (if  any)  for  given  values  of 
two  given  parts ;  and  therefore  in  addition  to  values  of  two 
parts  there  must  be  included  in  the  data  a  condition  by  which 
the  quadrant  of  any  one  of  the  three  unknowns  may  be  deter- 
mined. As  each  unknown  is  determined  by  a  function,  the 
algebraic  sign  of  any  other  function  (except  the  reciprocal 
function)  will  be  sufficient  to  determine  the  quadrant  of 
such  an  unknown. 

Given  a  =  86°  40',  c  =  32°40',  A  acute  ;  find  A  and  b. 


tan  A  =  tan  a  esc  c 

sec 

i  b  =  sec  a  sec  c 

1.2348 

1.2355 

0.2678 

0.0748 

1.5026 

1.3103 

J.  =  88°  12' 

6  =  87°  12' 

The  value  of  b  cannot  be  272°  48',  since  the  fourth  quad- 
rant is  inadmissible  by  either  of  the  two  formulae  : 

sin  b  =  sin  a  esc  A  and  tan  b  =  tan  c  sec  A. 

Had  A  been  known  to  be  greater  than  90°,  the  required 
values  would  have  been  A  =  268°  12'  and  b  =  272°  48'. 

Given  ^  =  213°,  6  =  54°  20',  and  C  known  to  be  greater 
than  180° ;  find  all  the  parts. 

tan  C  =  sec  b  cot  A  sin  a  =  sin  b  sin  A  tan  c  =  tan  b  cos  A 
0.2343                        9.9098  0.1441 

0.1875  9.7361  n  9.9236  n 

0.4218  9.6459  n  0.0677  n 

C  =  259°  16'  a  =  333°  44'  c  =  229"  27' 

Quadrant  given  by  sec  a  =  sec  A  sin  C,  sec  c  =  sin  ^  sec  0; 
or  tan  a  =  tan  b  cos  C,  sin  c  =  sin  6  sin  C. 


116 


SPHEKICAL  TRIANGLES. 


Given  h  =  53°  12',  C=  110°  48',  c  acute ;  find  a  and  A. 


tan  a  =  tan  b  cos  C 

0.1260 
9.5504  w 


tan  A=  secB  cot  C 

0.2226 
9.5796  n 


9.676471 
a  =  334°  37' 


9.8022  n 
^  =  327°  37' 


Given  A  =  327°  37',  0=  110°  48',  b  more  than  180°. 
sec  b  =  tan  A  tan  G    sec  a  =  sec  A  sin  (7    sec  c  =  sin  A  sec  C 

9.8022  n  0.0734  9.7288  n 

0.4204  n  9.9707  0.4496/1 

0.2226  =  0.0441  +  0.1784 

5  =  306°  48'  a  =  25°  23'  c  =  311°32' 

Given  ^  =  42°  20',  (7=  353°  50'.     Impossible.     Why? 
Given  ^  =  42°  20',    a  =  268°.  Impossible.     Why? 

Given  6  =  98°,  a  =  268°.  Impossible.     Why? 

Assume  the  values  of  any  two  parts  of  any  triangle  in 
any  row  of  the  following  table  and  compute  the  value  of 
any  other  part  in  that  row : 


A 

C 

a 

b 

c 

61°  35' 

30°  29' 

20° 17' 

23°  13' 

11°  32' 

61°  35' 

30°  29' 

339°  43' 

336° 47' 

348° 28' 

118°  25' 

329° 31' 

339°  43' 

336°  47' 

11°  32' 

298°  25' 

149° 31' 

339° 43' 

23°  13' 

11°  32' 

63°  15' 

135°  34' 

50°   0' 

120°  56' 

143°    5' 

296°  45' 

64°  26' 

310°    0' 

120°  56' 

143°    5' 

116°  45' 

224°  26' 

310°   0' 

239°   4' 

143°    5' 

63°  15' 

135°  34' 

310°   0' 

239°   4' 

216°  55' 

POLAR  AND  QUADRANTAL  TRIANGLES. 


117 


57.   Polar  triangles.  —  Let  ABC  be  a  spherical  triangle 
with  each  part  less  than  180°.     With  A,  B,  and  C,  respec- 


tively, as  poles,  describe  great  circles  B'C,  C'A',  and  A'B'. 
Now  if  with  A',  B',  and  C  as  poles,  great  circles  are  de- 
scribed, triangle  ABC  is  obtained.  The  two  triangles  ABC 
and  A'B'C,  thus  related,  are  said  to  be  polar  to  each  other. 
It  is  proved  in  geometry  that  the  relations  of  their  parts  are : 

a=lSO°-A';    b  =  lSO°-B';    c  =  180°-C". 

A  =  180°-a'',    B  =  180°-b';    C=180°-c'. 

Quadrantal  triangles.  —  If  the  triangle  ABC  is  a  right 
triangle,  its  polar  triangle  A'B'C  is  a  quadrantal  triangle. 
Hence,  substituting  in  (35),  (36),  and  (37) : 

sin  (180°  -  A')  =  sin  (180°  -B')  sin  (180°  -  a')  ; 
or  sin  A'  =  sin  B'  sin  a'.  (45) 

Similarly,  tan  ^'  =  sin  C  tan  a' ;  (46) 

tan  B'  =  —  tan  C  sec  a'.  (47) 

These  formulae  are  true  for  all  quadrantal  triangles  hav- 
ing each  part  less  than  180°. 


118  SPHERICAL  TRIANGLES. 

Now  assume  h,  Fig.  QQ,  to  be  a  quadrant. 

In  (45),  (46),  and  (47),  put  360°  -  A^  for  A\  180°  -  A  for 
J5',  180°  -  C,  for  C,  and  360°  -  a^  for  a'.  After  reduction 
there  is  obtained :  sin  A^ = sin  B^  sin  ai ;  tan  ^j = sin  (7i  tan  a ; 
tan  ^1  =  —  tan  Oi  sec  a^. 

Hence  the  formulae  (45),  (46),  (47),  are  true  for  a  quadran- 
tal  triangle  with  one  angle  greater  than  180°. 

The  triangle  made  up  of  the  hemisphere  +  the  triangle 
ABC,  Fig.  66,  has  its  angles  A,  180°  +  B,  and  180°  +  C,  and 
its  sides  h,  c,  and  360°  —  a.  These  values  also  satisfy  for- 
mulae (45),  (46),  (47).  The  values  of  the  parts  of  the 
triangle  AqBqCq,  Fig.  65,  also  satisfy  the  same  formulae. 
Hence  these  formulae  are  true  for  all  quadrantal  triangles. 
The  formulae  for  the  solution  of  a  quadrantal  triangle  are : 
(b  =  90°)  : 

sin  A  =  sinB  sin  a,  sin  C  =  sinB  sin  c, 

tan  A  =  sin  C  tan  a,  tan  C  =  sin  A  tan  c, 

sec  ^  =  sin  c  sec  a,  sec  (7  =  sin  a  sec  c, 

tan  B  =  —  tan  A  sec  c,  sec  JB  =  —  tan  a  tan  c, 

tan  B  =  —  tan  C  sec  a,  sec  5  =  —  sec  ^  sec  C. 

They  are  obtainable  from  the  formulae  for  right  triangles  by 
interchanging  the  large  and  small  letters  and  prefixing  a  minus 
sign  to  every  function  except  the  sine  and  cosecant. 

Isosceles  triangles  are  solved  by  drawing  a  perpendicular 
from  the  unequal  angle  upon  its  opposite  side,  thus  forming 
two  right  triangles. 

Regular  polygons  are  solved  by  joining  the  center  of  the 
polygon  to  each  vertex  and  also  to  the  middle  of  each  side, 
thus  forming  2  n  equal  right  triangles,  n  being  the  number 
of  sides. 


ELEMENTS  OF   SPHERICAL  FIGURES. 


119 


Elements  of  Some  Quadrantal  Triangles. 


a 

h 

c 

A 

B 

C 

240°   0' 

90° 

45°   0' 

225°   0' 

125° 16' 

144° 44' 

103°  26' 

90° 

154°   0' 

238°   0' 

240°  41' 

202°  28' 

292°  22' 

90° 

48°  50' 

300°  22' 

68°  55' 

135° 23' 

104°  53' 

90° 

133°  40' 

249° 12' 

255° 18' 

224°  24' 

174° 13' 

90° 

94°   8' 

175° 57' 

135° 34' 

135° 43' 

104° 53' 

90° 

133° 40' 

110°  48' 

104° 42' 

135°  36' 

299°  47' 

90° 

124° 51' 

307° 15' 

113° 29' 

48°  48' 

Elements  of  Some  Isosceles  Triangles. 


a 

h 

c 

A 

B 

C 

54°  20' 

54°  20' 

87°   5' 

47°   0' 

47°   0' 

116°   0' 

63°  b(j' 

63°  56' 

302°   0' 

105° 44' 

105° 44' 

294° 40' 

112° 48' 

112° 48' 

254°   9' 

56°  12' 

m''  12' 

240°   8' 

100°   4' 

100°   4' 

207°   2' 

42°  24' 

42°  24' 

198°   8' 

127° 50' 

127° 50' 

272°  52' 

42°  24' 

42°  24' 

238° 30' 

101°   7' 

101°   7' 

303°  31' 

83°  57' 

83° 57' 

302°  20' 

Elements  of  Some  Eegular  Polygons. 


n 

a 

C 

w 

a 

c 

3 

60° 

70°  32' 

9 

20° 

145° 11' 

4 

90° 

90°   0' 

10 

30° 

159° 53' 

4 

60° 

109°  28' 

12 

24° 

161°  52' 

3 

108° 

116°  34' 

15 

20° 

166°  40' 

5 

60° 

138°  11' 

18 

18° 

171° 14' 

6 

50° 

145°  42' 

20 

12° 

166°  34' 

8 

40° 

158° 57' 

30 

10° 

173°  22' 

120  SPHERICAL  TRIANGLES. 

OBLIQUE   SPHERICAL  TRIANGLES. 

58.  As  in  plane  triangles,  the  solution  is  effected  by 
dropping  a  perpendicular  from  a  vertex  upon  the  opposite 
side.  Two  right  triangles  are  thus  formed,  in  each  of  which 
a  value  of  the  perpendicular  is  found.  The  solution  is  then 
obtained  by  equating  the  values  of  the  perpendicular. 

This  method  fails  when  the  given  parts  are  the  three 
sides  or  the  three  angles. 

The  solutions  for  these  two  cases  will  now  be  found. 


Fig.  75. 

59.  The  in-  and  excircles.  —  Prolong  the  sides  AB  and  AC, 
and  bisect  the  angles  A,  B,  and  CBCi  by  great  circles. 
Prom  the  intersections  0  and  Oi  drop  perpendiculars  on  AB. 
With  0  and  Oi  as  poles,  describe  small  circles  with  radii 
OCq  and  OCi  respectively.  It  is  easy  to  see  that  the  various 
triangles  AOBq  and  AOCq,  AOiB^  and  AOiOi,  etc.,  are  equal 
in  pairs;  therefore  the  small  circles  touch  AC  in  Bq  and 
Bi,  and  BC  in  A^  and  A^. 

Then  exactly  as  in  plane  triangles  it  may  be  shown  that 
AGi  =  |(a  +  ^  4-  c)  =  s,  ACo  =s-d,  BCq  =  s  -  b,  BCi 
=  s  —  c,  etc. 

Since  the  angles  ABC -^  CBCi  =  180°,  the  sum  of  their 
halves,  or  OBO^  =  90° ;  therefore,  O^BC  =  90°  -  i  B. 


THE  IN-,  EX-,  AND  CIRCUMCIRCLES.  121 

In  triangle  AOiCi,   tan  ^  ^  =  esc  s  tan  r^. 

In  triangle  AOCq,         tan  r  =  sin  (s  —  a)  tan  ^  A.         (48) 

In  triangle  BO  Co,         tan  r  =  sin  (.s-  —  b)  tan  -J-  .B. 

In  triangle  BO^C^,       tan  ?*i  =  sin  (s  —  c)  cot  ^  B. 

Hence  tan  r  =  Vcsc  s  sin  (s  —  a)  sin  («  —  t)  sin  {s  —  c).  (49) 

60.  The  incircle  of  the  polar  triangle.  —  Since  (48)  is  true 
for  all  triangles,  it  is  true  for  the  polar  of  ABC.     That  is, 
tan  r'  =  sin  (s'  —  a')  tan  ^  A. 

Now         A'  =  180°  -  a  and  1  J['  =  90°  -  ^  a. 
2  s'  =  180°  -A-\- 180°  -  ^  + 180°  -  C;  hence  s'  =  270°  -  S, 
where  6'  stands  for    \{A-\-B+  C) ;    since    a'=  180°  — ^, 
s'  -  a'  =  90°  —  {S  —  A).     Substituting  these  values  in  (48) 
and  taking  the  reciprocal  of  the  resulting  equation, 

cot  r'  =  tan  \  a  sec  {S  —  A).  (50) 

The  same  process  applied  to  (49)  gives 

cot  r'  =  V-  cos  aS  sec  {S  -  A)  sec  {S  -  B)  sec  {S  -  C). 

The    circumcircle.  —  Bisect  C^ ^^^ 

the    sides    of    the    triangle  /'/^^^"'^^n^ ^"^X 

ABC      by      perpendiculars        //        \         ^^'.     \ 
meeting  in  F.     On  account     //  "^^        \         /    \^  \ 
of  the   equality  of  the  tri-    //         "^-.   ^^     '^  \  \ 

angles    in    pairs,   FA  =  FB   \J ^-% ^     \J 

=  FC,  which  denote  by  72.  ^K^  j  ~^^^ 

Also  the  angles  of  the  isos-    \    ^ ^^"^   I 

celes  triangles  are  equal  in      \  ^  / 

pairs.       Denote    FBC    and  \.  ^ 

FCB  by  X,  FCA  and  FAC  ^--^____^^-^'^ 

by  y,  FAB  and  FBA  by  z.  ^ig.  76. 

Since      x -{-  y  =  C,  y  +  z  =  A,  z-{-x  =  B,  x  =  S  —  A. 

In  triangle  BM^F,    tan  R  =  tan  \  a  sec  {S  —  A).  (51) 

By  (50),  tan  It  =  cot  r'.     Hence 

tan  i?  =  V-  cos  S  sec  {S  -  A)  sec  {S  -  5)  sec  (.S  -  C).    (52) 


122 


SPHERICAL  TRIANGLES. 


61.  Equations  (35)  to  (44)  applied  to  Fig.  77  give 

tan  C]  =  tan  h  cos  A  and  tan  Cg  =  tan  a  cos  B.  (53) 

tan  (7i  =  sec  h  cot  A  and  tan  Ca  =  sec  a  cot  JB.  (54) 

sin  p^suih  sin  J.   =  sin  a  sin  B.  (55) 

sin  j9  =  tan  Cj  cot  (7i  =  tan  c^  cot  Cg.  (56) 

tan  p  =  sin  Ci  tan  ^4  =  sin  c^  tan  ^.  (57) 

tan  p  =  tan  6  cos  C\  =  tan  a  cos  Cg.  (58) 

sec  p  =  sec  b  cos  q    =  sec  a  cos  Cg.  (59) 

sec  p  =  seGA  sin  d  =  sec  B  sin  Ca-  (60) 
Rearranging  (55)  and  changing  letters, 

sin  a  CSC  ^  =  sin  &  esc  jB    =  sin  c  esc  C.  (61) 

c 


62.   Methods  of  solving  oblique  spherical  triangles.  —  The 

following  methods  may  be  modified  in  various  ways  espe- 
cially to  suit  problems  in  which  all  of  the  unknown  parts 
are  not  desired. 

(1)  Given  two  sides  and  the  angle  opposite  one  of  them,  say 
a,  6,  A.    Find  B  by  (55),  c  by  (53),  and  C  by  (54)  or  (61). 


SOLUTIONS.  123 

(2)  Given  two  angles  and  the  side  opposite  one  of  them,  say 
A,  B,  h.     Find  a  by  {6b),  c  by  (53),  and  G  by  (54)  or  (61). 

(3)  Given  two  sides  and  the  included  angle,  say  h,  c,  A. 

First  find  c^  by  (53) ;  B  can  be  found  by  (57),  and  a  by  (59). 
If  C  is  desired,  it  can  be  found  by  (61).  When  only  one 
of  the  unknown  angles  is  desired,  draw  the  perpendicular 
through  the  angle  which  is  not  desired. 

(4)  Given  two  angles  and  the  included  side,  say  6,  A,  C. 
First  find  Gy  by  (54) ;  a  can  be  found  by  (58),  B  by  (60), 
c  by  (61).  If  only  one  of  the  unknown  sides  is  desired, 
draw  the  perpendicular  upon  the  side  which  is  not  desired. 

(5)  Given  the  three  sides.  —  The  angles  can  be  found  by 
(49),  (48),  etc. 

(6)  Given  the  three  angles.^ The  sides  can  be  found  by 

(52),  (51),  etc. 

SOLUTIONS. 

63.   1.   Given  two  sides  and  the  angle  opposite  one  of  them, 

a  =  236°5'2,  6  =  93°  40',  A  =  220%  and  B  known  to  be  be- 
tween  90°  and  180° ;  find  c  and  B. 

tan  Ci  =  tan  h  cos  A  cos  Cg  =  sec  h  cos  Cj  cos  a 

log  tan  6    1.1933/1  log  sec  6  1.1941  n 

Lgo^A     9.8843  n  Zcoscj  8.9211 

log  tan  ci  1.0776  Zcosa  9.7377  n 

Ci  =  85°  13',  c,  =  315°  27'.  L  cos  c^  9.8529 

sin  J5  =  sin  ^  esc  a  sin  h  Check :  tan  Cg  =  tan  a  cos  B 

L&mA  9.8081  n  Ztancg     9.9932  ?i 

log  CSC  a  0.0771  n  log  tan  a    0.1853 

isinft  9.9991  icos^     9.8081  n 

L  sin  B  9.8843  c  =  40°  40',  B  =  130°. 


124  SPHERICAL  TRIANGLES. 

2.  Given  two   angles  and  the  side  opposite  one  of   them, 
A  =  130°,  B  =  110°,  b  =  126°  58' ;  find  C  (less  than  90°). 

tan  Ci  =  sec  b  cot  A  sin  C2  =  sec  A  sin  Ci  cos  B 
log  sec  &    0.2209  n  log  sec  ^  0.1919 

icot^    9.9238  isinCi     9.9100 

log  tan  Ci  0.1447  L  cos  B     9.5341 

C  =  54°  22'  ^  ^^^  ^2     9.6360 

(72  =  25°  38',  and  hence  G  =  80°. 

3.  Given  two  sides  and  the  included  angle,   b  =  251°  30', 
c=275°  10',  A=132°  30',  and  a  less  than  180° ;  find  a  and  B. 

tan  Ci  =  tan  b  cos  ^ ;  Co  =  c  —  Ci ; 

sec  a  =  secb  cos  Cj  sec  c^ ;     tan  i?  =  tan  A  sin  c^  esc  Cg. 

log  tan  6    0.4755  log  sec  b   0.4985  ?i  log  tan  A  0.0379 

icos^     9.8297  n       Zcoscj     9.6470  ?i  i  sin  Ci  9.9524 

log  tan  Ci  0.3052  n       log  sec  Cg  0.0303  n  log  esc  Cg  0.4424 

log  sec  a  0.1758  ?2  log  tan  B  0.4327 

C2  =  158°50'  a  =131°  50'  5  =  290°  16' 

4.  Given  two  angles  and  the  included  side,  b  =  251°  30', 
A=132°  30',  (7=279°  50',  and  a  less  than  180° ;  find  a  and  B. 

tan  (7i  =  sec  5  cot  ^ ;  (72  =  (7  —  (7i ; 

tan  a  =  tan  b  cos  (7i  sec  C2 ;     sec  5  =  sec  ^  sin  (7i  esc  (72. 

log  sec  5    0.4985  n     log  tan  6    0.4755  log  sec  ^  0.1703  n 

icot^     9.9621  n     icos(7i     9.5148  n  LsinCi  9.9754 « 

log  tan  (7i  0.4606        log  sec  C^  0.0579  log  esc  C2  0.3153 

log  tang   0.0482 n  log  sec  ^  0.4610 

(72=    28°  56'  a  =  131°  50'  5  =290°  15' 


SOLUTIONS. 


125 


5.   Given  the  three  sides,  a  =  124°,   b  =  54°,   c  =  98°,  and 
A  less  than  180° ;     find  the  angles. 


tan  r  =  Vcsc  s  sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) ; 
^  ,    .      sin  (s  —  a)        ,  ,  „      sin  (s  —  6) .     .^ 


6.   Given  the  three  angles,  ^=220°,  ^  =  130°,  C  =  150°, 

and  a  known  to  be  more  than  180° ;  find  the  sides. 

tan R  =  V-  cos  5  sec (S  —  A)  sec  (S  —  B)  sec  {S  —  C); 

cot|a  =  ^^^,^^~^^  etc.      Check  by  (61). 
^  tani?      '  J  V    / 


Computation  for  Problem  5. 

Computation  for 

Problem  6. 

log  esc  138°     0.1745 

L(- cos  250' 

°)  9.5341 

i  sin  14°          9.3837 

log  sec  30° 

0.0625 

L  sin  84°         9.9976 

log  sec  120° 

0.3010  « 

L  sin  40°         9.8081 

log  sec  100° 

0.7603  n 

itan^r            9.3639 

logtan^i? 

0.6579 

L  tan  r            9.6819 

log  tan  R 

0.3289  n 

Jvcot^^         9.7018 

L  cot  ^  a 

9.7336  n 

log  cot  ^5       0.2957 

Lcot^b 

9.9721 

log  cot  iO       0.1062 

log  cot  ^  c 

0.4314 

^'=126°  34',   5  =  53°  42', 
and  0  =  76°  6'. 

If  A  is  known  to  be  greater 
than  180°,  the  negative  root 
of  tan^  r  would  have  been 
taken.    Hence  we  would  find 


a  =  236°  52',  6  =  93°  40', 

and  c  =  40°  38'. 


A  =  233°  26', 
G  =  283°  54'. 


B  =  306°  18' 


(See  Fig.  65.) 
As  a  check,  apply  (61). 


Zsina 
log  CSC  A 
Z/sin  b 
log  CSC  B 
L  sin  c 
log  CSC  C 

Check 


9.9229  n 

0.1919  n 

9.9991 

0.1157 

9.8137 

0.3010 

0.1148 


126  SPHERICAL  TRIANGLES. 

64.  It  is  frequently  convenient  to  express  a  desired  part 
directly  in  terms  of  given  parts,  especially  when  a  chain  of 
triangles  is  to  be  solved.  It  is  recommended  that  the  fol- 
lowing formulae  be  derived  from  a  diagram,  in  the  manner 
that  the  formulae  for  right  triangles  were  derived. 

They  will  here  be  obtained  by  applying  equations  (67) 
and  (68)  proved  in  Appendix  II.,  page  145. 

Thus,  from  (59),  (68),  and  (53), 

cos  a  =  cos  b  sec  Ci  cos  (c  —  Cj) 

=  cos  b  sec  Cj  (cos  c  cos  Cj  +  sin  c  sin  Cj) 
=  cos  b  cos  c  +  cos  b  sin  c  tan  Cj 
=  cos  b  cos  c+  cosb  sin  c  tan  b  cos  A ; 
cos  a  =  cos  b  cos  c  -f  sin  6  sin  c  cos  A. 
Multiply  by  esc  b  esc  c,  and  transpose  : 

cos  A  =  —  cot  b  cot  c  +  CSC  b  esc  c  cos  a. 
Apply  the  last  two  equations  to  the  polar  triangle : 
cos  A  =  —  cos  B  cos  C  +  sin  B  sin  C  cos  a, 
cos  a  =      cot  5  cot  (7  +  CSC  jB  csc  C  cos  A. 
Using  in  the  same  manner  (57),  (67),  and  (53), 
cot  jS  =  —  cos  c  cot  J.  +  sin  c  csc  A  cot  b, 
cot  b  =      cot  c  cos  A  +  csc  c  sin  A  cot  5. 

By  changing  letters,  similar  relations  between  other  parts 
of  the  triangle  may  be  obtained. 

The  application  of  these  forms  involves  the  use  of  both 
tables,  —  the  table  of  logarithms  of  functions  in  computing 
the  values  of  each  term  of  the  right-hand  member  in  any 
equation  and  the  table  of  natural  values  of  functions  in 
finding  the  answer. 

As  an  illustration  of  the  application  of  these  formulae, 
find  a  (less  than  180°)  given 

6  =  127°,         c  =  57°,        ^  =  130°. 


elp:ments  of  oblique  triangles. 


127 


cos  a  =  cos  b  cos  c  -f  sin  b  sin  c  cos  A. 


cos  6  cos  c  =  —  .3278 

sin  b  sin  c  cos  ^  =  —  .4305 
nat.  cos  a  =  —  .7583 

Therefore        a  =  139°  19' 


Lcosb 
Lcosc 

Lsinb 
L  sin  c 
i^  cos  A 


9.7795  n 
9.7361 
9.5156 11 
9.9023 
9.9236 
9.8081  n 
9.6340  n 


With  the  same  data,  find  B  (also  less  than  180°). 
cot  B  =  —  cos  c  cot  A  +  sin  c  esc  A  cot  h. 


.4570 

-  .8249 

-  .3679 

J5  =  110°12' 


L(-cosc)     9.7361)1 
Xcot^  9.9238  ?i 

9.6599 
L  sin  c  9.9236 

log  CSC  A        0.1157 
L  cot  6  9.8771  n 

9.9164  n 


Elements  of  Some  Oblique  Spherical  Triangles. 


a 

b 

c 

A 

B 

C 

131°  35' 

108°  30' 

84°  47' 

132° 14' 

110° 11' 

99°  42' 

124° 13' 

125° 42' 

82°  48' 

127°  22' 

128° 42' 

107° 33' 

134°  16' 

150°  57' 

55°  42' 

120°  48' 

144° 23' 

97°  43' 

126°  25' 

138°  32' 

45°  54' 

261°  16' 

234°  25' 

298°    6' 

309°  48' 

116°  45' 

129° 12' 

59°   4' 

274° 23' 

300°    5' 

120°  56' 

59°   4' 

106° 10' 

116° 45' 

63°  15' 

91°    7' 

113°   4' 

111° 41' 

79°  35' 

110°  50' 

109° 17' 

87°  35' 

126°   0 

152°   0' 

75°   0' 

142° 24' 

159°  16' 

133° 14' 

90°  53' 

117° 49' 

132°   5' 

120°   0' 

130°   0' 

140°    0' 

155°   5' 

147°    6' 

33°   2' 

133° 18' 

110° 10' 

70°  21' 

128  SPHERICAL  TRIANGLES. 

THE   CELESTIAL  SPHERE. 

65.  The  concave  surface  on  which  the  heavenly  bodies 
seem  to  be  situated  is  called  the  celestial  sphere. 

Intersections  of  the  celestial  sphere: 

I.  Its  intersection  with  the  plane  of  the  earth's  equator 
is  called  the  celestial  equator. 

II.  Its  intersections  with  the  prolongations  of  the  earth's 
axis  are  called  the  poles  of  the  heavens. 

III.  Its  intersection  with  a  plane  tangent  to  the  earth 
through  the  foot  of  an  observer  is  called  the  horizon  of  the 
observer. 

IV.  Its  intersections  with  the  vertical  line  through  an 
observer  are  called  the  zenith  and  nadir  respectively. 

Great  circles  through  the  poles  are  called  declination  cir- 
cles (or  hour  circles).  On  account  of  the  motion  of  the  earth 
on  its  axis,  these  circles  seem  to  move  from  east  to  west  15° 
every  hour. 

Great  circles  through  the  zenith  and  nadir  of  an  observer 
are  called  vertical  circles.  The  vertical  circle  through  the 
east  and  west  points  of  the  horizon  is  called  the  prime  verti- 
cal of  the  observer.  The  vertical  circle  through  the  north 
and  south  points  passes  also  through  the  poles ;  it  is  called 
the  meridian  of  the  observer. 

The  points  of  intersection  of  the  celestial  equator  with 
the  path  of  the  sun  are  called  equinoxes.  The  vernal  equi- 
nox is  passed  by  the  sun  on  March  21,  and  the  autumnal 
equinox  is  passed  on  September  21. 

The  arc  of  the  celestial  equator  from  the  vernal  equinox 
to  the  declination  circle  through  a  heavenly  body,  say  a  star, 
is  called  the  right  ascension  of  the  star.  The  right  ascen- 
sion is  usually  expressed  in  terms  of  time  (15°= 1  hour)  and 
is  measured  toward  the  east.     The  right  ascension  of  the 


THE   CELESTIAL  SPHERE. 


129 


meridian  at  any  time  is  given  by  a  sidereal  clock  set  for  that 
meridian. 

The  angle  which  the  declination  circle  makes  with  the 
meridian  (=  arc  on  the  equator  from  meridian  to  declination 
circle)  is  called  the  hour-angle  of  the  star.  The  arc  of  the 
declination  circle  from  the  celestial  equator  to  the  star  is 
called  the  declination  of  the  star.  The  declination  is  positive 
when  measured  from  the  equator  to  the  north. 

The  arc  of  the  horizon  from  its  south  point  to  the  vertical 
circle  through  a  star  is  called  the  azimuth  of  the  star.  The 
arc  of  the  vertical  circle  from  the  horizon  to  the  star  is 
called  the  altitude  of  the  star.  The  azimuth  is  positive 
when  measured  in  the  direction  SWNE,  and  the  altitude  is 
positive  when  measured  toward  the  zenith. 

These  definitions  are  illustrated  in  the  following  diagram. 


N  north 

P  pole 

Z  zenith 

V  equinox 

S  star 


QZ 

EQW 

SiWNE    horizon 


VD  right  ascension 

QPD  hour  angle 

DS  declination 


EZW        prime  vertical       SiWNA    azimuth 


NPZSi      meridian 


AS 


altitude 


The  triangle  PSZ  is  sometimes  called  the  Astronomical 
Triangle.     Its  parts  are  : 

Arc  PZ=  arc  QP—  arc  QZ=  90°  -  latitude. 
Arc  ZS  =  arc  AZ  -  arc  ^.S'  =  90°  -  altitude. 
Arc  SP  -  arc  DP  -  arc  DS  =  90°  -  declination. 


130  SPHERICAL  TRIANGLES. 

Angle  PZS  =  arc  S^  WNA  -  arc  S,  WN  =  azimuth  -  180°. 
Angle  ZSF  is  called  the  parallactic  angle  of  the  star. 
Angle  ZPS  =  arc  QD  =  hour-angle. 

Notation.  —  The  hour-angle  will  be  denoted  by  P,  the 
declination  by  d,  the  azimuth  by  Z,  the  altitude  by  a,  the 
latitude  by  I. 

APPLICATIONS  TO   GEOGRAPHY. 

115.  Find  in  geographical  miles  (60  miles  =  1°)  the 
shortest  distance  on  the  surface  of  the  earth  from  East- 
port  (10°  E.  longitude,  44°  50'  N.  latitude)  to  Point  Barrow 
(80°  W.  longitude,  71°  20'  N.  latitude). 

116.  Find  the  shortest  distance  on  the  surface  of  the 
earth  from  Quito  (2°  W.  longitude,  0°  latitude)  to  Point 
Barrow. 

117.  A  required  point  is  described  as  being  equidistant 
from  Eastport,  Point  Barrow,  and  Quito.  What  is  the 
distance  on  the  surface  of  the  earth  from  the  point  to  each 
of  the  three  places  ? 

118.  Find  the  distance  from  Eastport  to  San  Diego 
(40°  W.,  32°  40'  N.). 

119.  The  latitude  of  Key  West  is  24°  10'  N. ;  the  latitude 
of  Attou  Island  is  53°  N.  The  distance  apart  of  the  two 
places  is  3722  geographical  miles.  Find  their  difference  in 
time. 

120.  A  ship  sails  from  Halifax  (lat.  44°  40'  N.,  long.  13° 
25'  E.)  in  a  direction  N.  81°  E.,  and  continues  on  an  arc  of  a 
great  circle  for  2472  miles.     Find  the  place  arrived  at. 

APPLICATIONS  TO  ASTRONOMY. 

121.  Given  the  latitude  of  a  place  and  the  declination 
of  the  sun,  to  find  the  azimuth  of  the  sun  at  sunrise  and 
the  length  of  the  day. 


APPLICATIONS.  181 

Consider  the  following  cases : 
i.   I  =  40°,  d  =  -\-23°27';  the  longest  day. 
ii.   I  =  40°,  d  =  0° ;  days  and  nights  equal. 

iii.   Z  =  40°,  d=-23°27';  the  shortest  day. 

iv.   I  =  66°  33',  d  =  23°27';  or  I  and  d  complementary. 

122.  Given  the  latitude  of  a  place  and  the  declination 
of  the  sun,  to  find  when  the  sun  is  due  west  (in  the  prime 
vertical).     Consider  the  following  cases : 

i.   I  =  40°,  d  =  +  23°  27. 
11.   Z  =  40°,  (Z  =  0°. 
ill.   Z  =  66°33',  (Z  =  4-23°27'. 

123.  Given  the  latitude  of  a  place  and  the  declination 
of  the  sun,  to  find  the  altitude  and  azimuth  of  the  sun  at 
6  o'clock  in  the  morning.     Take  i  =  40°,  d  =  23°27'. 

124.  The  interval  which  elapsed  between  the  transits  of 
Kegulus  over  the  prime  vertical  was  observed  to  be  4  hours 
and  12  minutes.  Find  the  latitude  of  the  place,  the  decli- 
nation of  Regulus  being  12°  40'. 

125.  At  about  8  o'clock,  the  altitude  of  Regulus  was 
observed  to  be  16°  40'.  Three  hours  and  twenty-four  min- 
utes later,  it  was  again  observed  and  found  to  be  40°  20'. 
Find  the  latitude  of  the  place. 

126.  The  interval  between  the  observed  equal  altitudes 
of  Antares  was  4  hours  and  20  minutes.  Find  the  lati- 
tude of  the  place  if  the  observed  altitudes  were  54°  20' 
and  the  declination  of  Antares  is  —  26°  6'. 

127.  From  a  place  north  of  the  equator,  the  altitude  of 
Capella  was  observed  to  be  76°  40'  when  its  hour-angle  was 
1  hour  and  6  minutes,  the  declination  of  Capella  being 
45°  54'.     Find  the  latitude  of  the  place  of  observation. 


132  SPHERICAL  TRIANGLES. 

A  tree  100  feet  high  stands  on  a  horizontal  plane  in  40^^ 
north  latitude.  What  is  the  area  described  by  its  shadow 
between  the  hours  of  9  and  2  on  the  21st  of  March  ? 

Let  h  be  the  height  of  the  tree,  s  the  length  of  its 
shadow,  Z  the  azimuth  of  the  sun  and  of  the  shadow,  P 
the  hour-angle,  I  the  latitude,  and  a  the  altitude  of  the  sun. 

First,  cos  Z  =  —  tan  a  tan  I,  and  by  plane  trigonometry, 
s  tan  a  =  h.  The  product  of  these  is  s  cos  Z  =  —  h  tan  I. 
But  —  s  cos  Z  is  the  northerly  projection  of  the  shadow,  and 
h  tan  Z  is  a  constant.  Hence,  on  this  day  the  amount  which 
the  shadow  extends  north  is  a  constant.  Therefore  the  path 
of  the  shadow  of  the  top  of  the  tree  is  a  straight  line,  due 
east  and  w^est,  83.91  feet  from  the  foot  of  the  tree. 

Second,  cot  Z  =  sin  I  cot  P,  from  which  the  azimuth  of  the 
shadow  at  9  o'clock  is  237°  16'  and  at  2  o'clock  it  is  138°  4'. 

The  area  of  the  triangle  passed  over  by  the  shadow  be- 
tween 9  and  12  is  5477  square  feet,  and  the  area  described 
between  12  and  2  is  3162  square  feet,  making  the  total  area 
8639  square  feet. 


AREA  OF  A   SPHERICAL  TRIANGLE. 

66.   Denote  the  area  by  K,  the  half  sum  of  the  angles  by 
S,  the  radius  of  the  sphere  by  R.     Then,  by  geometry, 

90° 
Thus,  if  ^  =  130°,  B  =  110°,  C  =  80° ;  K=  2.4435  R\ 
Also,  if  ^  =  220°,  B  =  130°,  C  =  150° ;  K=  5.5851  R\ 

128.  Given  h  =  70°  20',  c  =  38°  28',  A  =  52°  30' ;  find  K. 

129.  Given  a  =  139°  21',  Z>  =  12G°58',  c  =  56°52';  find  i^T. 


APPENDIX   I. 


THEORY  OF  LOGARITHMS. 


67.  Arithmetic  and  geometric  series.  —  Logarithms  are  a 
series  of  numbers  severally  assigned  to  ordinary  numbers, 
and  so  chosen  that  the  multiplication  and  division  of  or- 
dinary numbers  may  be  accomplished  by  the  addition  or 
subtraction  of  their  logarithms. 

Thus,  if  the  ordinary  numbers  are  terms  in  a  geometric 
progression  (a  multiplication  series)  and  to  each  term  we 
assign  a  term  in  an  arithmetic  progression  (an  addition 
series),  it  can  be  shown  that  the  multiplication  or  division 
of  the  terms  in  the  geometric  progression  can  be  accom- 
plished by  the  addition  or  subtraction  of  terms  in  the 
arithmetic  progression. 


Thus  suppose  it 
is  desired  to  multi- 
ply the  numbers 
9-^(32)  and  i^  (S). 
The  logarithm  of 
1^ (32)  isa  +  5d  (5), 
and  the  logarithm 
oi^^(8)isa+7d(7). 
Adding  these  loga- 
rithms gives  2  a 
+  12rf(12),  which 
is  larger  than  the 
logarithm     of     the 

133 


Numbers. 

Logs. 

,,10 

a 

r« 

a-\-       d 

,.8 

«+    2rf 

,.7 

a+    3rf 

r« 

a-f    4rf 

1^ 

a-f    5d 

,A 

a-h    6rt 

7-3 

a+    Id 

9-2 

a+    8c? 

r 

a-f    9(? 

1 

a  + 10  d 

Numbers. 

Logs. 

1024 

0 

512 

1 

256 

2 

128 

3 

64 

4 

32 

5 

16 

6 

8 

7 

4 

8 

2 

9 

1 

10 

134  THEORY  OF  LOGARITHMS. 

correct  product  [?•«  (256)]  by  a  +  10  d  (10). 
But  a  +  lOd  (10)  is  the  logarithm  of  1. 
Hence      log  (r^  x  t^)  =  log  r^  +  log  r'  —  log  1, 
log  (32  X  8)  =  log  32  +  log  8  -  log  1. 

And  this  relation  is  perfectly  general  in  any  two  such 
series.  Either  by  trial  or  reduction  from  this  relation  we 
may  obtain 

log  (pq)  =  \ogp-\-  log  q  -  log  1, 

log  ^  =  logp  -  log  g  +  log  1, 

logp"  =  n  logp  —  (n  —  1)  log  1, 

log  -Vp  =  -(logp  +  log  1). 
n 

It  will  be  seen  that  a  logarithm  is  a  function  fulfilling 
the  condition 

Function  (ab)  =  function  (a)  H-  function  (6)  —  a  constant. 

The  theory  of  logarithms  is  principally  devoted  to  ob- 
taining the  logarithm  of  any  number  whatever,  —  in  other 
words,  interpolating  in  the  two  series. 

Logarithms  may  be  considered  geometrically,  as  they 
were  by  the  inventor  of  the  first  system: 

NAPIERIAN    LOGARITHMS.* 

68.   Suppose  two  points,     Ao      ^        A 
A  and  B,  to   start  at  the 
same   time   with   the   same 
initial   velocity.      A   starts 
at  Ao  and  moves  along  the 


1-^^ — '^ 4 


£ — k 


Fig.  79. 
half -line  A^A^  with  a  uniform  velocity.     B  starts  at  B„  and 

*  This  name  has  by  some  writers  been  incorrectly  apphed  to 
Speidell's  Logarithms,  which  were  published  in  1619,  five  years  after 
the  publication  of  Napier's  logarithmic  tables. 


NAPIKUIAN   LOGARITHMS.  135 

moves  along  the  segment  -B^B^,  with  a  changing  velocity 
such  that,  when  B  arrives  at  any  point  U^,  the  velocity  of  B 
will  be  proportional  to  the  remaining  segment  B^B^.  Then, 
if  at  any  time  A  is  at  A^  and  at  the  same  time  B  is  at  B^, 
the  length  of  A^A^  was  called  by  Napier  the  logarithm  of 

The  numerical  value  of  the  length  of  B^B,  was  taken  by 
Napier  to  be  10',  or  10,000,000.  It  will  here  be  denoted 
by  k ;  hence  from  the  definition,  Napierian  log  k  =  zero. 

Suppose  that  during  the  first  unit  of  time  A  moves  a  dis- 
tance 1,  and  B  moves  a  distance  which  denote  by  (1  —  r)fc.t 
The  distance  of  the  moving  point  B  from  B,  will  then  be 
k  —{1  —r)k  =  rk  ;  and  hence  Napierian  log  ?'A;  =  1. 

Now  during  the  second  unit  of  time  A  will  move  another 
distance  equal  to  1,  making  its  distance  from  A„  equal  to  2. 
Since  the  law  of  the  diminution  of  the  velocity  of  B  remains 
the  same  throughout  its  motion,  we  obtain  the  distance 
traversed  by  it  in  the  second  unit  of  time  in  the  manner  as 
it  was  obtained  for  the  first  unit  of  time.  That  is,  multiply 
the  distance  which  it  is  away  from  B^  at  the  beginning  of 
the  unit  of  time  under  consideration  by  the  constant  1  —r. 
This  gives  for  the  space  traveled  over  by  B  in  the  second 
unit  of  time  (1  —  r)  rk ;  so  that  B  is  then  removed  from  B^ 
by  the  distance  rk—{l—  r) rk  =  r^k. 

Hence  Napierian  log  i^k  =  2. 

Similarly,  Napierian  log  r^k  =  3 ; 

and  in  general,        Napierian  log  r"A:  =  n. 

*  Mirifici  Logarithmornm  Canonis  Constructio.  English  transla- 
tion by  W.   R.  Macdonald,  Edinburgh,  1889. 

t  If  the  initial  velocities  of  A  and  B  are  equal  as  assumed  by 
Napier,  and  k  is  very  large,  the  value  of  r  will  be  nearly  equal,  but 
slightly  less  than  1.  The  distance  gone  by  B  in  the  first  unit  of  time 
will  be  slightly  less  than  1  on  account  of  the  retardation  of  the  velocity 
of  ^  as  it  approaches  Bg. 


136  THEORY  OF  LOGARITHMS. 

It  will  be  seen  that  the  numbers  form  a  geometric  pro- 
gression and  their  logarithms  form  an  arithmetic  progression, 
and  hence  that  Napierian  logarithms  are  only  a  particular 
case  of  the  more  general  logarithms.  Napier  solved  the 
problem  of  finding  corresponding  intermediate  terms  in 
the  two  series ;  his  logarithms  are  no  longer  used,  but 
the  method  by  which  we  could  interpolate  in  the  Napierian 
series  is  quite  similar  to  the  method  adopted  for  the  same 
process  applied  to  systems  now  in  use. 

PRESENT  SYSTEM  OF  LOGARITHMS. 

69.   Suppose  two  points  A  and  B,  moving  in  the  same  direc- 
tion, A  with  a  uniform  ve- 
locity along  the  line  A„A,,   1"  "'    t" ■^-^— 

and  B  along   the  half-line     . I 

^„^,,  with  a  velocity  directly    £  £      '^^      ET  '+^~" 

proportional  to  its  distance  ^^"     " 

from  Bo-  If  at  any  time  the  distance  of  A  from  A„  is  A^^, 
and  at  the  same  time  the  distance  of  B  from  B„  is  B^B^, 
the  length  A„A^  is  defined  as  the  logarithm  of  B^B^. 

If  it  is  assumed  that  A  is  at  zero  (yl„)  when  J5  is  at  1  (Sj), 
the  logarithm  of  1  =  0.  This  assumption  avoids  the  neces- 
sity of  subtracting  the  logarithm  of  1  when  multiplying  by 
means  of  logarithms. 

Denote  the  velocity  which  B  has  acquired  when  at  B^  by 
V ;  then  at  any  other  point  the  velocity  of  B  will  be  v  times 
its  distance  from  B„ ;  thus,  the  velocity  of  B  at  any  point 
B^  =  v  times  B„B^.  In  the  first  unit  after  B  leaves  B^,  A 
will  be  supposed  to  move  a  distance  1,  and  B  will  move  in 
the  same  time  a  distance,  which  may  be  denoted  by  cv  times 
B„B^]  for  a  particular  system  of  logarithms,  c  and  v  are 
constants. 

Let  us  count  from  the  instant  that  A  was  at  A„  and  B  at 
^1.     At  this  instant  B^  will  coincide  with  Bi,  and  the  dis- 


NATURAL  LOGARITHMS.  137 

tance  of  B  from  B„  =  1.  In  the  first  unit  of  time  afterward, 
the  distance  moved  over  by  B  becomes  cv,  and  its  total  dis- 
tance from  B^  at  the  end  of  this  first  unit  of  time  is  1  -\-  cv\ 
therefore,  log(l  +  cv)=  1. 

Under  the  same  assumptions,  in  the  second  unit  of  time 
A  moves  another  unit.  This  makes  the  distance  A„A,  =  2, 
and  the  motion  of  ^  =  cv  times  its  distance  from  B^  at  the 
beginning  of  this  unit  of  time  =  cv{\  +  cv).  Therefore,  at 
the  end  of  the  unit  of  time,  the  distance  of  B  from  B^ 

=  1  -f  cv  +  cv  (1  +  cv)  =  (1  +  cvy. 

Therefore  log  (1  +  cvf  =  2. 

By  a  similar  process  of  reasoning  it  may  be  shown  that 
log  (1  +  cvy  =  3,  and  in  general  that  log  (1  +  cvy  =  n. 

70.  It  will  be  seen  that  the  values  of  the  distances  of 
A  from  A„  at  the  ends  of  successive  units  of  time ;  namely, 
1,  2,  3  •••  n,  form  an  arithmetic  progression:  and  that  the 
values  of  the  distances  of  B  from  B„\  namely,  (1  -\-cv), 
{l-\-cvy,  (1  -\-cvy  '"  (l+cv)",  form  a  geometric  progres- 
sion. The  value  of  c  depends  on  the  value  assumed  for  y;,  but 

V  is  perfectly  arbitrary,  and  values  for  it  may  be  assumed 
at  pleasure ;  now  a  system  of  logarithms  may  be  constructed 
by  assuming  for  v  any  value  at  all,  since  it  has  already 
been  shown  that  the  terms  of  any  arithmetic  progression 
may  be  used  as  logarithms  of  the  terms  of  any  geometric 
progression.  But  in  systems  of  logarithms  in  actual  use, 
special  values  have  been  given  to  v  on  account  of  special 
conveniences  served  thereby.  Thus,  in  higher  mathematics, 
the  reciprocal  of  v  arises  as  a  factor ;  this  factor  (denoted 
by  M)  is  eliminated  and  processes  simplified  by  making 
M=  1  (whence  also  v  =  1).     The  system  in  which  M  and 

V  are  taken  as  unity  is  called  the  Natural  System  of  Loga- 
rithms, and  the  logarithms  in  that  system  are  called 
Natural  Logarithms,  or  Speidell's  logarithms. 


138  THEORY  OF  LOGARITHMS. 

71.  The  number  corresponding  to  a  given  logarithm. 

Let  us  now  imagine : 

(1)  That  B,  having  an  initial  velocity  of  v,  moves  for  one 
unit  of  time  divided  into  n  parts. 

(2)  That  B,  having  an  initial  velocity  of  1,  moves  for  v 
units  of  time,  each  unit  being  divided  into  n  parts. 

In  each  case  denote  the  value  of  the  constant,  which  is  c 

for  one  unit  of  time,  by  —for  each  -  of  the  unit  of  time, 
m  n 

Then   under   the   first   assumption  the  motion  of   B  is 

1  +  —  )  ;  and  under  the  second  assumption  it  is  ( 1  4-  — 
mj  \       mj 

By  the  binomial  formula, 

l  +  !iY^l  +  gH  +  »(»-l)f  +  "(»-l)(n-2)^Vetc. 
mJ  m        1  '2  '  m^  1  •  2  •  3  •  m^ 

-,    ,    IV"     1    ,  ^v  ,  n^v^  —  nv  ,  nV  — 3nV-f-2nv  .     , 

1  +  -     =1+  —  +  - — ^r -„H .     „    „   ^„ hetc. 

mJ  m      1  '  2  '  m^  1  •  2  •  3  •  m^ 

Now  when  n  is  very  great,  the  distance  that  B,  with  a 
velocity  v  (or  1  under  the  second  assumption)  will  go  from  B^ 

in  -  of  a  unit  of  time  will  approach  as  a  limit  -  f  or  -  ].    But 
n  .  n\      nj 

this  distance  is  also  v  (or  1)  times  — .     Therefore,  when  ri  is 

m 

infinite,  m  =  n.     Also,  as  m  and  n  become  very  great, 
nin-^  and  J^^  both  approach  ^    ""^     : 

n(n-l)(n-2)v^  ^^^  nh^-3n^v^+2nv  ^  _nV 

1.2.3.m3  1.2.3.m3        ^^  1.2.3.m^ 

When  these  limiting  values  are  put  in  the  values  of  the 
motions,  these  values  of  the  motions  become  identical. 
Denote  the  common  result  by  b,  which  is  either  the  motion  of 
B  from  Bi  for  one  unit  of  time  with  an  initial  velocity  of  /,  or  the 


BASE   OF  THE   NATURAL  SYSTEM.  139 

motion  of  B  from  B^  for  v  units  of  time  with  an  initial  velocity  of  i. 

The  value  of  h  is 

In  this  value  of  &,  make  v  =  1  and  denote  the  result  by  e ; 
e  is,  therefore,  the  motion  of  B  from  B^  while  A  goes  a  dis- 
tance equal  to  1,  both  ^  and  5  having  the  same  initial 
velocity.     The  value  of  e  is 

The  numerical  value  of  e,  which  is  readily  computed,  is 
2.71828  +  .     (See  page  10.) 

After  B  was  at  By  (or  1)  its  motion  for  a  unit  of  time  with 
an  initial  velocity  of  1  made  its  distance  from  B„  equal  to  e; 
hence  another  unit  of  time  will  make  its  distance  1  x  e  x  e=e^ ; 
and  at  the  end  of  v  units  of  time  the  distance  of  B  from  B„ 
will  be  e" ;  but  this  =  6,  the  distance  that  B  would  be  from 
jBo  at  the  end  of  one  unit  of  time,  if  its  velocity  when  at  5, 
was  V.  Therefore  at  the  end  of  the  second  unit  of  time, 
with  an  initial  velocity  of  -y,  the  distance  of  B  from  Bq 

=  1  X  6  X  5  =  6^  =  1  X  e"  X  e"  =  e^" ; 

and  for  n  units  of  time  with  an  initial  velocity  of  v,  the  dis- 
tance of  B  from  B^=  5"  =  e"";  and  hence  from  the  defini- 
tion, 6"  =  e""  is  the  number  whose  logarithm  is  n.  That  is, 
if  ^  is  a  number  and  n  is  its  logarithm, 

6"  =  N,  where  h  =  e*. 

Taking  two  numbers,  Ni  and  N^,  with  logarithms  Wj  and  n2, 

iVi  X  iVg  =  6"!  X  5"*  =  6''^+"«. 

n 

JV'  =  6""  and  v^=  b\ 


140  THEORY  OF  LOGARITHMS. 

That  is,   log  (N^  x  N^)  =  log  Ni  +  log  iVg, 
log  (N,  -  N,)  =  log  N,  -  log  N,, 
log  N^  =p  log  iV, 
log  -y/Nzzz  log  JV-7-  r. 
It  will  be  noticed  that 
The  numbers  form  a  geometric  progression ; 
Their  respective  logarithms  form  an  arithmetic  progression; 
These  systems  of  logarithms  are  only  particular  cases  of 

the  more  general  form  of  logarithms ; 
Logarithms  in  these  systems  satisfy  the  condition : 
function  {ah)  =  function  {a)  -f-  function  (h)  ; 
The  constant  (log  1)  which  was  introduced  in  the  Napierian 
system  will  be  zero  in  these  systems. 

72.  Base.  —  In  the  relation  6"  =  N,  we  see  that  the  loga- 
rithm n  is  that  exponent  of  a  number  h  corresponding  to  a 
number  N.  This  is  expressed  by  saying  that  in  these 
systems,  a  logarithm  is  an  exponent ;  the  number  h  which  is 
affected  by  this  exponent  is  called  the  base  of  the  system 
of  logarithms.  It  is  evident  that  if  N  remains  constant  and 
h  is  arbitrarily  made  to  change,  then  n  will  also  change. 

Relations  between  logarithms  in  systems  with  different 
bases.  —  Let  x  be  the  logarithm  of  a  in  a  system  whose  base 
is  h  (written  x  =  log,,  a),  and  z  be  the  logarithm  of  &  in  a 
system  whose  base  is  a  (written  z  =  log,  b).  Then  h''  =  a 
and  a"  =  b. 

Now  in  any  of  the  present  systems,  taking  the  logarithms 
of  each  side  of  these  equations, 

X  log  b  =  log  a ;  z  log  a  =  log  b ;  whence  xz  =  1. 

That  is,  log6  a  x  loga  6  =  1.  (63) 

Hence  if  the  logarithm  of  a  number  to  any  base  is  known, 
its  logarithm  to  any  other  base  can  easily  be  found. 


THE  LOGARITHMIC  SERIES.  141 

THE    LOGARITHM    CORRESPONDING    TO    A    GIVEN 
NUMBER. 

73.  We  have  heretofore  regarded  the  logarithms  as  in- 
creasing uniformly,  and  the  rate  of  increase  of  the  numbers 
continually  changing.  We  will  now  find  a  relation  giving 
the  increase  in  the  logarithm  when  the  numbers  increase 
uniformly. 

Now  equation  (62)*  gives  the  value  oi  b  =  e",  and  hence 
putting  nv  for  v, 

e-.  =  6"  =  l  +  „.+^_  +  ___  +  etc., 

which  is  the  number  whose  logarithm  is  n.  Denote  this 
number  by  1  +  s ;  then 

s  =  nv  -\-  - — -  + Y  etc. 

1.21.2.3 

Assume  now  that  n  =  As-\-  Bs^  +  Cs^  +  etc.,  and  substitute 
this  assumed  value  of  n  in  the  value  of  s;  and  let  us  for 
the  present,  in  squaring  and  cubing  the  assumed  value  of 
n,  drop  powers  of  s  higher  than  the  third.  Equating  the 
coefficients  of  the  like  powers  of  s,  there  is  obtained 

Av  =  1,   whence  A  =  -\ 

V 

Bv+^  =  0,   whence  JB  =  --^; 

Cv  +  ABv^+  —  =  0,   whence  C=  —  ; 
6  3  V 

and  the  law  of  the  series  being  seen,  it  is  unnecessary  to 
repeat  the  operation  retaining  higher  powers  of  s.     Hence, 

„  =  log(l  +  .)=l[.-|V|-lVf-etc.]. 

Changing  the  sign  of  s, 


142  THEORY   OF   LOGARITHMS. 

Subtracting  these  two  equations,  and  remembering  that 
the  subtraction  of  logarithms  corresponds  to  a  division  of 
numbers, 

p  —  q  1.  -\-  s      p 

Now  let  s  =  ;  whence  a  _    =  — 

-f =l?^^Kf^^Kfi|)■-«■}  (". 

If  g  =  1  and  7?  =  5,  log  5  =  1.     Substitute  and  solve  for  v : 
o[b-l  ,  Ifb-iy  ,  1/b-iy  ,     ,   "I 

whence  v  can  be  computed  for  any  positive  base. 

Note  that  if  the  base  is  1,  v  is  zero ;  and  that  if  the  base 
is  less  than  1,  v  is  negative. 

Modulus.  —  The  reciprocal  of  v  is  called  the  modulus  of 
the  system  of  logarithms.  Now  v  has  been  shown  to  be  the 
ratio  of  the  rate  of  increase  in  the  numbers  in  a  system 
whose  base  is  b  (A  at  A^,  B  at  B^  to  the  rate  of  increase  in 
the  numbers  in  the  natural  system,  the  logarithms  being 
conceived  to  increase  uniformly.  Hence  the  reciprocal 
of  V,  or  the  modulus  of  the  system  whose  base  is  b,  is  the 
ratio  of  rate  of  increase  in  the  logarithms  in  the  system 
whose  base  is  b  {A  at  A^,  B  at  ^1)  to  the  rate  of  increase  in 
the  logarithms  in  the  natural  system,  the  numbers  being 
conceived  to  increase  uniformly. 

It  will  be  seen  from  (64)  that  a  logarithm  in  any  system 
is  inversely  proportional  to  the  value  of  v  for  that  system. 
Now  V  is  the  reciprocal  of  M,  and  hence  a  logarithm  in  any 
system  is  directly  proportional  to  the  modulus  of  the  sys- 
tem. The  modulus  of  the  natural  system  is  1 ;  the  logarithms 
of  a  number  in  any  other  system  can  therefore  be  obtained  by 
multiplying  the  natural  logarithm  of  the  number  by  the  modu- 
lus of  the  system. 


THE   CURVE   OF  LOGAUITHMS. 


143 


If  in  (64)  M  is  written  instead  of  -,  and  logq  is  trans- 
posed, we  obtain 

From  which  the  logarithm  of  any  positive  number  can  be 
found  when  the  logarithm  of  any  other  number  and  also 
the  modulus  of  the  system  are  known. 

74.   The  logarithmic  curve.  —  In  the  following  diagram  the 
abscissae  of  points  in  the  curve  represent  numbers,  and  the 
corresponding  ordinates  represent  the  natural  logarithms  ot 
these  numbers. 
Y 


Fig.  81. 


75.  Solve  the  following  equations.  —  After  obtaining  the 
general  solution,  substitute  2  for  a,  3  for,  b  4  for  c,  5  for  d, 
6  for  e,  7  for/;  obtain  the  numerical  answer  in  each  case. 

130.  a**+«  =  (Z«+^.  ^^^     ^a'+f^b, 

131.  a*V'  =  e. 

132.  ab'  =  cd'\ 
i     a'  =  lf>, 


(  a'+y  =  b, 
\  h^-y  =  a. 


135.   a^'b'^c  =  1. 


133. 


( (f+<i  =  e^+^. 


136.    a''  =  c. 


APPENDIX  II. 


1.   GONIOMETRY.      2.    COMPLEX   QUANTITIES. 
3.   HYPERBOLIC  FUNCTIONS. 

In  the  subject  of  Goiiiometry  we  will  consider :  (1)  func- 
tions of  the  sum  or  ditt'erence  of  angles ;  (2)  the  circular 
measure  of  an  angle ;  (3)  arc  functions. 

Review  Chapter  I.     Functions  of  Angles. 

76.   Functions  of  the  sum  or  difference  of  angles. 

Denote  the  value  of  angle  BAG  by  ic,  the  value  of  angle 
CAD  by  y.  From  any  point  D 
in  AD  drop  a  perpendicular  on 
AG  and  on  AB.  From  G  draw- 
parallels  to  BA  and  DE,  respec- 
tively. 

The  angles  GDF  and  BAG 
are  equal  because  their  sides 
are  respectively  perpendicular. 

AD  sin  (x  +  y) 
=  ED  =  BG-\-FD. 

But       BG=  AG  sinx 
and  AG  =  AD  cos  y. 

Also  FD  =  GD  cos  x  and  GD  =  AD  sin  y. 

Substituting :  sin  (x  +/)  =  sin  x  cosy  +  cos  x  sin/.       (65) 

Also,  AD  cos  (x-\-y)=AE  =  AB-  FG, 

AB  =  AG  cos  X  =  AD  cos  x  cos  y, 

FG  =  GD  sin  x  =  AD  sin  x  sin  y. 

Hence  cos  (x  +y)  =  cos  x  cosy  —  sin  x  sin/. 

144 


Fig.  82. 


(66) 


SUM  OR  DIFFERENCE  OF  ANGLES.  145 

Had  the  figure  been  drawn  with  either  x  or  y  in  any- 
other  quadrant  than  the  first,  the  same  result  would  have 
been  obtained. 

Had  y  been  subtracted  from  x,  the  point  F  would  have 
been  to  the  right  of  O;  FC  would  have  been  drawn  from 
F  to  the  left,  and  FD  would  have  been  drawn  from  F  down- 
ward. 

Hence       sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y.  (67) 

Also  cos  (x  —  y)  =  cos  x  cos  y  -f  sin  x  sin  ?/.  (68) 

,  .       sin  (x  +  y)      sin  x  cos  y  -f  cos  x  sin  y 

tan  (x  +  ?/)  = ) =-f  = ; ,— ^-. 

^      cos  (x  -\-  y)      cos  X  cos  y  —  sin  a;  sin  y 

Multiply  numerator  and  denominator  of  this  equation  by 
sec  X  sec  y.     There  is  obtained 

tan  X  -\-  tan  y 


tan  {x  +  y) 
Similarly,        tan  {x  —  y) 


1  —  tan  X  tan  y 
tan  X  —  tan  y 


1  -|-  tan  a*  tan  y 

The  other  functions  of  (x  +  ?/)  and  of  (x  —  ^^)  can  be  ob- 
tained by  taking  the  reciprocals  of  sin  (x  ±  y),  cos  (x  ±  y), 
and  tan  {x  ±  y). 

By  adding  and  subtracting  (65)  to  (68), 

sin  X  cos  y  =  ^  [sin  (x  -{-y)  -{-  sin  (x  —  ?/)],  (69) 

cos xsmy  =  ^  [sin  (x -\- y)  —  sin  (x*  —  ?/)],  (70) 

sin  X  sin  ?/  =  —  J  [cos  (a?  +  ?/)  —  cos  (a;  —  y)'] ,  (71) 

cos  a;  cos  y  =  \  [cos  (a;  +  ?/)  +  cos  (a;  —  ?/)].  (72) 

In  these  relations  put  ^(A-\-B)  for  x  and  i(^— JB)  for  ?/. 

sin  ^  +  sin  JB  =  2  sin  1  (^  +  B)  cos  ^  (^  -  B),  (73) 

sin  ^  -  sin  B  =  2  cos  L  (^  -j-  5)  sin  i  (.4  -  ^),  (74) 

cos  ^  -  cos  i  =  -  2  sin  ^(A+  B)  sin  J-  (^  -  75),  (75) 

•  cos  A  +  cosB  =  2  cos  ^(A^  B)  cos  J  (.1  -  5),  (76) 


146  GONIOMETRY. 

TRIGONOMETRIC  EQUATIONS. 

77.   1.    Solve  the  equation  a  sina^  -f  h  cos  x=  c. 

Put  for  cos  X  its  value  Vl  —  sin^ic,  and  solve  for  sin  x : 

sm  X  =  — ^ —^ 

a- 4-  ¥ 

Or  proceed  thus :,  Assume  an  angle  y  such  that  h  =  a  tan?/. 
Substitute  in  the  given  equation  and  multiply  by  cos  y : 

a  (sin  X  cos  y  +  cos  a;  sin  y)  =  G  cos  ?/. 

The  solution  is  therefore  obtained  by  obtaining  y  from 

the  equation  tan  y  =  -  and  then  finding  x  from  the  equation 
a 

sin  (x  -\-y)  =-  cos  2/. 

The  first  method  of  solution  has  the  advantage  of  ex- 
pressing sin  X  directly  in  terms  of  the  given  parts ;  the 
second  method  has  the  advantage  of  using  formulae  adapted 
to  logarithmic  computation. 

2.   Solve  the  equation  a  tan  x-{-b  cot  x  =  c. 
Multiply  by  tan  x  and  arrange : 

a  tan^  x  ~  c  tan  x-\-  b  =  0. 

From  which  quadratic  equation 


,  c  ±  Vc^  —  4  a6 

tan  X  = 

2a 

Or  proceed  thus :  from  (11)  and  (9) 

1  —  cos  2 .'»       1      ,         1  +  cos  2  X 

tan  X  = r-^ and  cot  x  =  — '-^— 

sm  2x  sm  2  a? 

Substitute  in  the  given  equation  and  arrange: 

c  sin  2  X  -f-  (a  —  b)  cos  2  a;  =  a  +  ^• 

^„  ,  a  —  b       .     ,„      ,     .       a  +  b 

If  tan  y  = ,    sm  (2x-^y)  = cos  y. 

G  C 


TKIG(JN()METRIC   EQUATIONS.  147 

Tins  solution  i,nves  a  loj^aritlimic  solution  of  the  general 
equation  of  the  second  degree.  Thus  in  az^  —  cz-\-b  =  ^ 
divide  by  2,  and  put  tan  x  for  z. 

The  equation         a  tan  x  -f  h  cot  x=^c  is  obtained. 

Hence  z  can  be  found  by  lirst  finding  ?/,  then  2x-\-y, 
then  X ;  finally,  z  =  tan  x. 

3.  Solve  a  (sin  x  -f  tan  x)  =  h  sin  x  (esc  a;  +  cot  x). 
This  equation  can  be  arranged  thus : 

(cos  ic  +  1)  (a  sin  x  —  h  cos  a;)  =  0, 

and  the  roots  are  obtainable  by  equating  each  factor  sepa- 
rately to  zero. 

4.  Solve  the  equation  a  sin  x  -{-h  sin  {x  -f- 10°)  =  c. 
By  applying  (Qto)  and  arranging,  this  becomes 

{a-\-b  cos  10°)  sin  x-\-h  sin  10°  cos  x=iCy 

which  is  of  the  form  a  sin  x-\-h  cos  x  =  c. 

5.  Solve  the  equation  sin  x  +  sin  {x  —  10°)  =  .6. 

In  (73),  write  x  instead  of  A  and  a;  —  10°  instead  of  B\ 
there  is  obtained 

sin  X  +  sin  (x  —  10°)  =  2  sin  {x  —  5°)  cos  5°. 

Comparing  with  the  given  equation, 

sin  (a; -5°)  =  .3  sec  5°;  hence  a;  =  22°  32'  or  167°  28'. 

6.  Solve  the  equation  sin  x  sin  {x  —  10°)  =  .6. 

In  (71),  put  X  —  10°  for  y  and  compare  the  resulting  equa- 
tion with  the  given  equation 

cos  (2  a;  -  10°)  =  cos  10°  -  1.2  =  -  .2152, 

aj  =  56°13'  or  133°  47'. 


148  GONIOMETEY. 

7.  Solve  the  equation   sin  x=  .6  sin  (x  —  10°). 
Apply  (67), 

sin  x=  .6  sin  x  cos  10°  —  .6  cos  x  sin  10°. 

Multiply  by  esc  10°  esc  x ;  there  is  obtained 

.6  cot  a;  =  .6  cot  10°  -  esc  10°. 
From  which  x  =  135°  43'  or  315°  43'. 

8.  Solve  the  equation   tan  x  tan  (x  —  10°)  =  .6. 

The  quotient  of  (71)  and  (72)  compared  with  the  given 
equation  gives 

cos  (2  a; -10°) -cos  10°^  ^ 
cos  (2  a; -10°)+ cos  10° 

which  gives         cos  (2x  —  10°)  =  .25  cos  10°, 

whence  x  ==  42°  52'  or  147°  8'. 

Problems.  —  Find  the  least  positive  angle  satisfying  each 
of  the  following  equations : 

137.  sin  (10°  +  a;)  =  .6  sin  (20°  +  x). 

138.  tan  a;  =  .6  tan  (a;  —  10°). 

139.  tan  {x  -h  10°)  +  tan  (x  -  10°)  =  2  cot  x. 

140.  sin  {x  +  10°)  +  cos  (a;  -  10°)  =  cos  (a;  -|-  10°). 

141.  cos  X  cos  (x  — 10°)  =  .6. 

142.  32  cos^  a;  +  3  tan  2  a;  +  3  cot  (45°  +  a;)  =  0. 

143.  sin  (20°  +  x)  cos  (20°  -  a;)  =  .6. 

144.  sin  2  a;  +  sin  3  a;  =  3  sin  x. 

145.  sin^  x  -  sin2  {x  -  70°)  =  .6. 

146.  cot  X  tan  2  a;  —  tan  x  cot  2  a;  =  2. 

147.  sin  (c  -f  x)  CSC  (c  —  x)  =  cos  {a  +  b)  sec  (a  —  6). 


CIRCULAR  MEASURE.  149 

78.  The  circular  measure  of  an   angle  — Let  A  be  the 

center  of  a  circle,  and  CB  q^ 

the  arc  included  between 

the  sides  of  angle  A=0.* 

With    the    same    center 

A   and   any   radius   ABi 

draw  an  arc  BiCi  also  in-  _^ 

eluded  between  the  sides  ' 

of  A  Fio.83. 

The  sectors  ABC  and  AByCi  are  similar,  and  hence 
arc  BC :  radius  AB  =  arc  BiCi ;  radius  ABi. 
That  is,  the  ratio  arc:  radius   is  constant   for   any  angle. 
This  ratio  is  taken  as  the  measure  of  angles ;  its  magnitude 
for  any  angle  is  called  the  circular  measure  of  that  angle. 
That  is,  in  circular  measure : 

1  arc 

angle  = 


radius 

The  arc  of  an  angle  generated  by  half  a  revolution  (180°) 
is  T  times  the  radius.     Hence 

angle  of  half  a  revolution  =  tt. 
The  unit  angle  is  obtained  by  dividing  this  equation  by  tr. 
unit  angle  =  ^^g^^  ^-^  ^^^^  ^  revolutjon^ 

TT 

This  unit  angle  has  been  called  a  radian.     Its  value  in 

degrees  can  be  obtained  from  this  last  equation, 

180° 
1  radian  =  ^^^  =  57.296°  =  57°  17'.75. 

TT 

If  the  angle  0  in  circular  measure  has  x°, 

arc  of  angle  0         degrees  in  arc  of  angle  6 


radius  degrees  in  arc  equal  to  radius 

That  is,  (9  =  —^-^  and  a;  =  57.296  d. 

*  Angles  measured  in  radians  will  have  their  values  denoted  by 
Greek  letters,  w  (Pi),  0  (Theta),  </>  (Phi),  etc. 


150 


GONIOMETRY. 


A  table  for  converting  degrees  into  radians,  or  for  con- 
verting radians  into  degrees,  will  be  found  at  the  bottom 
of  page  vii  of  the  tables. 

Exercises.  —  The  following  table  gives  the  values  of 
various  angles,  first,  in  degrees  and  minutes ;  second,  in 
radians ;  third,  in  terms  of  an  angle  of  half  a  revolution : 


Angle  in 

Angle  in 

Angle  in 

Angle  in 

Angle  in 

Angle  in 

Degrees. 

Radians. 

Terms  of  tt. 

Dk(;kees. 

Radians. 

Terms  of  tt. 

172° 

3.0020 

.9556 

144°  9' 

2.5133 

.8008 

100° 

1.7453 

M56 

169°  23' 

2.9563 

.9401 

125° 

2.1566 

.6944 

104° 51' 

1.8300 

•  .5825 

179° 

3.1241 

.9944 

204°  12' 

3.5639 

1.1344 

194° 

3.3859 

1.0778 

195°  33' 

3.4130 

1.0864 

300° 

5.2360 

1.6667 

274°  10' 

4.7851 

1.5231 

344° 

6.0039 

1.9111 

289°  50' 

5.0585 

1.6102 

354° 

6.1784 

1.9667 

344°  19' 

6.3355 

2.9667 

79.  Arc  functions.  —  The  equation  ^=sin^  gives  the 
value  of  A^  as  a  function  of  6;  it  is  frequently  convenient 
to  express  the  value  of  ^  as  a  function  of  A^.  The  notation 
employed  is :  6  =  arc  sin  N*  This  will  be  read,  6  is  the 
angle  (or  arc)  whose  sine  is  JSf.  Similarly  0  =  arc  tan  N  is 
read,  6  is  the  angle  (or  arc)  whose  tangent  is  N;  and  simi- 
larly for  each  of  the  other  functions. 

It  will  be  seen  that  if  we  omit  the  name  of  a  trigono- 
metric function  prefixed  to  one  side  of  an  equation,  the 
name  of  the  corresponding  arc  function  is  prefixed  to  the 
other  side  of  the  equation ;  and  reversely. 

It  must  be  remembered,  however,  that  in  the  equation 
A'=  sin^,  there  is  but  one  value  for  N  for  each  value  of  6; 


*  By  some  authors  this  is  written,  d  =  sin-i  iV. 


ARC   FUNCTIONS.  151 

but  in  the  equation  $  =  arc  sin  N,  there  are  an  infinite  num- 
ber of  values  of  6  for  any  value  of  the  number  N.  Thus 
in  the  equation  N=  sinO,  if  0  =  ^tt  or  30*',  N'=  M;  but  in 
the  equation  d  =  arc  sin  N,  if  N=  .5,  0=[n-{-  (  —  l)"^-]7r, 
n  being  any  whole  number. 

Exercises.  —  1.    Find  6  =  arc  tan  1. 
Here  tan  0  =  1,  and  0  =  (n  -{-  ^)7r. 

2.  Find  N=  tan  (arc  cos  ^). 

Let  arc  cos ^  =  0,  then  cos  6  =  ^;  6  =  (2n  ±  ^)7r. 
Hence  N=  tan  (2n  ±  i)7r  =  ±  1.7321. 

3.  Prove  that        sin(arc  sin  N)  =  N, 
and  that  arc  sin  (sin  0)  =  6. 

2  N 

4.  Prove  that        2  arc  tan  N=  arc  tan- -• 

1  —  -^^ 

5.  Arc  sin  N  -\-  arc  sin  -J-  -^  =  \  tt.     Find  N. 

6.  Arc  sin  2  iV—  arc  sin  iV  V3  =  arc  sin  N.     Find  N. 

PROBLEMS  IN  GONIOMETRY. 

148.  At  a  distance  of  70  feet  from  the  foot  of  a  tower, 
the  angle  of  elevation  of  the  top  is  21°  greater  than  the 
angle  of  elevation  seen  at  a  distance  of  160  feet  from  the 
foot  of  the  tower.     Find  the  height  of  the  tower. 

149.  At  a  distance  of  296  feet  (a)  from  the  base  of  a 
tower,  the  angle  of  elevation  of  the  top  is  three  times  as 
great  as  the  angle  of  elevation  seen  from  a  point  208  feet 
(6)  nearer  the  base.     Find  the  height  of  the  tower. 

150.  The  wind  strikes  a  horseman  at  an  angle  of  50°  with 
the  road  when  he  is  traveling  at  the  rate  of  8  miles  an  hour ; 
and  at  an  angle  of  40°  when  his  rate  is  12  miles  an  hour. 
Find  the  direction  and  velocity  of  the  wind, 


152  COMPLEX   QUANTITIES. 

151.  From  the  top  of  a  hill  360  feet  above  the  surround- 
ing plain,  an  officer  sees  a  column  of  the  enemy's  infantry- 
advancing  directly  toward  the  hill.  The  column  is  known 
to  be  one  mile  long,  and  at  a  certain  time  the  whole  column 
is  seen  by  the  officer  to  subtend  an  angle  of  2°  7'.  At  that 
time  what  is  the  distance  of  the  front  rank  from  the  middle 
of  the  base  of  the  hill  ? 

152.  A  circle  has  a  radius  of  20  inches  (r).  From  the 
same  point  in  the  circumference  two  chords  are  drawn  of 
lengths  10  inches  (ki)  and  15  inches  (A^g),  respectively.  Find 
the  area  of  the  portion  of  the  circle  bounded  by  these  chords 
and  by  the  arc  joining  the  ends  of  the  chords.  (Two  solu- 
tions.) 

153.  A  circular  field,  surrounded  by  a  fence,  contains  1 
acre.  A  cow  is  fastened  to  the  fence  by  a  chain,  which, 
when  measured,  was  found  to  be  55.53  feet  long.  Over 
what  area  can  the  cow  graze  inside  the  circle? 

COMPLEX   QUANTITIES. 

80.  A  complex  quantity  is  of  the  form  a  -\-  hi,  where  i 
stands  for  V—  1 ;  a  is  called  the  real  part  of  the  complex 
quantity,  and  hi  is  called  the  imaginary  part. 

It  has  already  been  explained  that  the  multiplication  of 
a  segment  by  —  1  reverses  the  direction  of  the  segment. 
This  reversal  may  be  conceived  as  being  made  by  means  of 
two  half-reversals.  Since  i'^  =  —  l,  i  will  be  taken  as  the 
multiplier  which  first  turns  OU  (Fig.  84)  into  OUx,  and  then 
turns  0  Ui  into  0  U^ ;  multiplying  again  by  i,  0  U^.  is  turned 
into  OlT'a;  and  another  multiplication  by  i  turns  0X1^  into 
OU. 

Real  positive  quantities  have  been  conceived  as  laid  hori- 
zontally to  the  right,  and  real  negative  quantities  to  the 
left.     Positive   imaginary  quantities  will   be  conceived  as 


ARGUMENT  AND   MODULUS. 


153 


being  laid  vertically  upward,  and  negative  imaginaries  ver- 
tically downward. 

Thus  the  complex  quantity  a-\-bi  will  be  represented  by 
laying  off  0M=  a  to  the  right  of  the  origin  and  MAi  =  h 
upward  from  M.  If  a  is  negative  it  is  laid  off  to  the  left 
of  0,  and  if  b  is  negative  it  is  laid  off  downward. 


Fia.  84. 


Denote  the  angle  MOA^  by  0  and  the  length  of  the  line 
OAi  by   m.     Then   a  =  h  tan  $  =  m  cos  0   and   6  =  m  sin  0. 

Hence  the  complex  quantity  a  -f  bi  may  be  written : 
m  (cos  6  -\-  i  sin  0).  The  angle  0  is  called  the  argument  of 
the  complex  quantity,  and  m  is  called  its  modulus. 

It  will  be  noticed  that  the  segment  OU"  (+  1)  is  changed 
into  the  segment  OAi  by  two  distinct  operations. 

First.  A  turning  oi  OU  through  the  argument  6  into 
OUa'  This  turning  is  therefore  equivalent  to  a  multiplica- 
tion of  + 1  by  cos  0  -\-i  sin  6. 

Second.  A  multiplication  of  the  length  of  OUa  by  the 
niodulus  m,  the  positive  square  root  of  a^  +  b^. 


154 


COMPLEX   QUANTITIES. 


Fig.  85. 


Addition  of  complex  quantities.  —  Suppose  it  is  desired  to 
add  the  complex  quantities  mi  (cos  ^  4-  **  sin  B)  and  m^  (cos  <^ 
-f- 1  sin  <^).  Make  angle  M0A=6,  OA=m^ ;  angle  NAB=(fi, 
AB=m2',  the  required  sum  is  OA-\-AB.  The  sum  of  the  real 
parts  is  m^  cos  6  -\-  m^,  cos  <^ 
{0M-\-  AN)]  and  the  sum 
of  the  imaginary  parts  is 
i  (mi  sin  6  -\-  wig  sin  <^) ;  that  is, 
MA  +  NB. 

Had  mg  (cos  <^  +  i  sin  <l>) 
=  OC  been  laid  off  first  and 
mi  (cos  <i>  +  i  sin  </>)  =  CB  been 
added  to  it,  the  result  would 
have  been  precisely  the  same. 
Hence  the  segment  OB  rep- 
resents the  sum  both  in  mag- 
nitude and  direction, 

A  third  complex  quantity  can  be  added  to  this  sum  by 
drawing  from  B  a  segment,  say  BD,  representing  in  magni- 
tude and  direction  this  third  complex  quantity.  Then  the 
segment  OD  will  represent 
both  in  magnitude  and  direc- 
tion the  sum  of  the  three  com- 
plex quantities,  and  so  on. 

Subtraction.  —  To     con- 
struct the  difference  of 
ma  (cos  ^  +  ?*  sin  <;^) 
and    mi  (cos  $  -\-  i  sin  $), 
lay  off  the  minuend  (OA); 
let  AB  represent  the   sub- 
trahend;   lay    it    off    back- 
ward from  AB  to  the  point 

jBj  ;  then  OB^  will  represent  the  remainder  both  in  magni- 
tude and  direction. 


FUNDAMENTAL  OPERATIONS. 


156 


Multiplication,  —  The  product  of  the  two  complex  quanti- 
ties m]  (cos  6  4- 1  sin  6)  and  mg  (cos  <f>  -{- i  sin  <^)  is 

7iiim2  [(cos  0  cos  (^  —  sin  ^  sin  <t>)  -{-i  (sin  $  cos  <^  +  cos  $  sin  <j!))] 
=  miTTia  [cos  (^  +  <^)  +  i  sin  (^  -f  <^)],      (77) 

the  arguments  being  added  and  the  moduli  multiplied. 

eTust  as  OU  (+1)  was 
turned  tlirough  an  angle  0 
and  then  multiplied  by  m 
when  operated  upon  by  the 
multiplier 

m  (cos  6  -\-i  sin  6), 
so  OA  is  turned  thl-ough 
an  angle  <t>  (UOB  ov  AOC) 
and  then  multiplied  by  m^ 
when  operated  upon  by  the 
multiplier 

m2{Gos<f)  +  ism<t>). 

If  (Fig.  87)  angle  UOA  = 
e,  UOB  =  <f>,  UOC=e  +  <l>,  then  BOC=UOA.  Also  the 
length  OC  of  the  modulus  =  m^m^  is  a  fourth  proportional 
to  1,  ma,  and  mi;  that  is,  OU:  0B=  OA:  OC.  Hence  the 
two  triangles  UOA  and  BOC  are  similar. 

To  construct  OC,  therefore,  it  is  sufficient  to  make  angle 
BOC  =  e  and  angle  B  =  angle  OUA. 

If  now  OC  is  operated  upon  by  a  new  multiplier, 

m2{cos\l/-\-ismif/)y 

the  new  argument  will  be  6  +  cf)-{-if/,  and  the  new  modulus 
will  be  m-^tn^m^'  The  geometric  construction  can  be  obtained 
as  before. 

Division.  —  Rationalizing  the  denominator  of  the  fraction 

7)iy  (cos  6  -\-i  sin  6) 
'w?2(cos  ^-f-i  sin  <^) 


Fig.  87. 


156 


COMPLEX   QUANTITIES. 


gives  ^  (cos  ^  +  I  sin  6)  (cos  <f>  —  i  sin  <^) 

=  ^[cos  ^  +  ^  sin  ^][cos  (-  <^)  +  i  sin  (-  <^)] 

=  ^  [cos  {e~^)  +  i  sin  (0  -  <^)], 
the  arguments  being  subtracted  and  the  moduli  divided. 

A 


To  construct  the  quotient  OC,  make  angle  AOC=  <f>  and 
angle  A  =  B.     Then 

the  argument  UOC=  UOA  -  UOB  =  6-<t>', 

and  since    OC :  0U=  OA :  OB,  the  length  of  00  = 

the  modulus  of  the  quotient  =  — ^' 

Powers  and  roots.  —  The  factor  m  (cos  6  -\-i  sin  B)  repeated 
n  times  gives  by  (77) : 

[m  (cos  ^  -h  i  sin  ^)]'*  =  m"(cos  nO  -h  *  sin  nO).        (78) 

n 

Write  -  instead  of  6,  m  instead  of  m",  and  take  the  nth 
n 

root 

-  /       fi  f)\  .         - 

m"  (  cos  -  4- 1  sin  -  )  =  [m  (cos  0-\-i  sin  $)Y. 
\      n  nj 


POWERS   AND   ROOTS. 


157 


Fia.  89. 


To  construct  the  power,  lay  off,  upward  from  the  half-line 
OU,n  angles  each  equal  to  UOAi  =  6. 

Make    OAi  —  m    and    join    UA^.        ""^^ 
Make    angles    OA^A^,    OAiA^,   etc.,  ^^^^ 

each  equal  to  OCMi-     Then  0C7=  1, 
OAi  =  m,  OA2  =  m^,  OA3  =  m*,  etc. 

Now  the  arguments  UOA^,  UOA^f 
UOA^y  etc.,  increase  in  arithmetical 
progression ;  the  segments  OAi,  OA2, 
OA^,  etc.,  which  represent  the  com- 
plex quantities  m  (cos  $  +  i  sin  0), 
m^  (cos  d-\-  i  sin  Oy,  m^  (cos  6-\-i  sin  fff, 
etc.,  increase  in  geometrical  pro- 
gression. 

It  follows  that  the  arguments  $, 
2  6,  3  0,  etc.,  are  the  logarithms  of 

the  corresponding  complex  quantities.  Now  denote  m  by 
k^  and  take  the  ^h  root  of  k^  (cos  6  +  i  sin  6).  This  root  is 
k  (cos  1  +  I  sin  1),  which  is  the  number  whose  logarithm  is  1, 
and  therefore  the  base  of  this  system  of  logarithms. 

The  curve  UA1A2AS  (Fig.  j^ 

89)  is  called  the  logarithmic 

spiral.     It  will  be  seen  that 

it  may  be  used  as  a  graphical     /  /     •  \  ;^ 

table     of    logarithms ;     the 

same  thing  is   true   of  the  1  i^:^ lU 

curve  in  Fig.  81. 

When  m  =  1,  the  points 
Ai,  A2,  As,  etc.,  will  be  ar- 
ranged at  equal  distances 
from  each  other  on  a  circle 
of  unit  radius.  Fig.  90.     In 

this  case  k  also  equals  1  and  the  base  of  the  system  of 
logarithms  is  cos  1  +  i  sin  1  =  .5403  -f  .8415  i. 


158 


COMPLEX   QUANTITIES. 


Construction  of  roots.  —  It  has 
already  been  shown  that  the  nth 
root  of  m  (cos  0  -{-i  sin  d)  was 

mn(  cos--f  ism- ), 
\      n  nj 

where  m  is  the  positive  square     r 

root  of  a^  +  6^ ;  m»  will  here  de- 
note the  real  positive  nth  root  of 
m.  Now  the  functions  of  2  stt 
+  0  are  the  same  as  those  of  0, 
and  hence 


m  (cos  0  +  i  sin 


6)  \n  =  ran 


cos 


7i  n 

and  by  taking  values  of  s  =  0,  1,  2,  3,  etc.,  the  n  roots  are 

1  /       ^ 
m"(  COS-  +  I  sin- 


m"  (  cos f- 1  sm 


2,7  + 

71 


Taking  ^  =  y5_^  ^  50°^  m  =  3.052,  ti  =  5,  the  roots  are 
1.25  (cos    10° +  1  sin   10°),      1.25  (cos   82°  +  ^  sin   82°), 
1.25  (cos  154°  +  i  sin  154°),      1.25  (cos  226°  -f  i  sin  226°), 
1.25  (cos  298° +  ^  sin  298°), 
giving  the  points  E^,  R21  ^^3?  -^45  ^s- 

By  taking  m=l  and  x=0,  the  71  roots  of  +1  are  obtained. 
By  taking  m=l  and  x=tt,  the  71  roots  of  —1  are  obtained. 
By  taking  771= 1  and  a;=|7r,  the  n  roots  of  i  are  obtained. 
By  taking  m=l  and  a;=  |  tt,  the  n  roots  of  —i  are  obtained. 


SERIES   FOR  8IN0  AND   COS  0.  I59 

81.  Values  of  functions  of  multiple  angles.  —  Put  1  for  m 
in  (78)  :         cos  nO  +  i  sin  nO  =  (cos  6  -\-  i  sin  Oy.  (79) 

Expand  by  the  binomial  formula ;  equate  the  imaginary 
parts  and  also  the  real  parts : 

sin  ne  =  n  cos^-^^  sin  0  -  ^(^-^K^-^)  cos"-^^  sin^  $  -f-  etc. 

COS  n6  =  cos"^  -  ^^r""-^)  cos'*-^^  sin^  9  +  etc. 
1  •  ^ 

Thus  to  obtain  the  functions  of  3  6,  make  w  =  3, 

sin  3  (9  =  3  cos^  ^  sin  ^  -  sin^  0  =  3  sin  0-4:  sin»  0,    (80) 
cos  3  ^  =  cos^  i9  -  3  cos  ^  sin^  0  =  4:  cos^  ^  -  3  cos  0. 

82.  Exponential  value  of  cos  9  +  /  sin  6.  —  When  0  is  zero, 
cos  0  i-i  sin  0is  1.  Now  let  us  imagine  a  point  generating 
a  circle ;  when  the  angle  is  zero,  counted  from  a  horizontal 
initial  line,  the  horizontal  component  of  the  motion  of  the 
generating  point  is  zero,  and  the  vertical  component  of  its 
motion  is  equal  to  the  motion  in  the  arc.  Therefore  when 
0  is  at  zero,  the  rate  of  increase  of  cos^  +  i  sin  0  is  i  times 
the  rate  of  the  generating  point.  If  we  assume  the  gener- 
ating point  to  move  a  unit  distance  in  unit  time,  the  rate  of 
increase  (0  =  zero)  of  cos  ^  + 1  sin  ^  is  i. 

Now  0  is  the  logarithm  of  cos  0  -\-ism0;  therefore,  by 
the  theory  of  logarithms,  cos  0  +  i  sin  0  =  e^,  v  being  the 
rate  of  increase  in  the  number  cos  0  -i-i  sin  0,  when  0  is 
zero  and  increasing  at  unit  rate. 

Therefore  cos  0  -\-ism0  =  e^ ; 

Changing  the  sign  of  0 :  cos  ^  —  ?'  sin  ^  =  e~^; 

whence  sin  0  =  — ^  (e^*"  —  e~^') 

_e  0^       ,  0" 0'  .       .^.. 

~1      1.2.3'^1.2.3.4.5      1.2.3.4.5.6.7^         ^    ^ 

cos(9  =  i(e«*  +  e-^') 


160 


HYPERBOLIC  FUNCTIONS. 


HYPERBOLIC  FUNCTIONS. 

83.  Definition  of  an  equilateral  hyperbola.  —  Suppose  a 
double  cone  (called  a  cone  of  two  nappes)  with  vertex  at 
V.     If  this  cone  is  cut 

through   the    axis,   two     -  ^ 

Singles,  BVE  Siud  DVC, 
will  be  formed.  If  the 
cone  is  such  that  this 
angle  is  90°,  and  the 
cone  is  again  cut  by  a 
plane  OA  normal  to  the 
plane  BVE,  this  new 
section  is  bounded  by 
a  curve  of  two  branches 
QAR  and  Q^A^Ri;  this 
curve  is  called  an  equi- 
lateral hyperbola.  The 
points  A  and  Ai  are 
called  the  vertices  of  the 
hyperbola,  the  point  O, 
midway  between  them, 
is  called  the  center  of 
the  hyperbola,  and  AA^ 
is  called  the  axis  of  the 
hyperbola. 

The  two  following 
properties  of  the  equi- 
lateral hyperbola  are 
proved  in  treatises  deal- 
ing with  this  curve. 

First.   If  Q  is  a  point  on  the  curve,  and  O^is  the  abscissa 

ofQ, 

ON'  -  NC/  =  0A\ 


Fig.  92. 


Fig.  93. 


THE   EQUILATERAL   HYPERBOLA. 


161 


Second.   Area  segment  AQN  oi  the  hyperbola 


^ONxNQ 


iOA' 


'-■Fin- 

Now  the  sector  OAQ  =  triangle  ONQ  —  segment  AQN. 
Hence  sector  OAq  =  \  OAHogJ^^^J^^\ 

The  quantity  log 


ON 


^  ]  will  be  denoted  by  u:  and 
OA      J  *^     ' 

the  semi-axis  OA  will  be  denoted  by  a. 
Hence  Sector  OAQ  =  la^u. 

84.  Definitions  of  the  hyperbolic  functions.  —  Let  0  be  the 
center  of  an  equilateral  hyperbola,  0A(=  a)  its  semi-axis, 
Q  any  point  on  the  curve, 
ON  the  abscissa  of  Q,  and 
NQ  the  ordinate.  With  a 
center  0  and  a  radius  a, 
describe  a  circumference; 
let  P  be  a  point  on  the  cir- 
cumference ;  let  the  length 
of  arc  AP  be  equal  to  aO ; 
then  6  represents  the  cir- 
cular measure  of  the  angle 
AOP.  The  area  of  the  cir- 
cular sector  AOP  equals  its  arc  times  ^  the  radius  =  ^a^O. 

*  1.  The  equation  of  the  two  generators  of  the  cone  are  x—y=0  and 
X  +  y  =  0,  their  product  being  x'^—y'^=0.  As  these  generators  revolve 
around  the  axis  of  X,  x  remains  constant  and  y^  becomes  y^  +  z'^. 
Hence  the  equation  of  the  double  cone  is  x'^  —  y^  —  z^  =  0.  If  now  the 
cone  is  cut  by  the  plane  z=  a,  the  projection  of  the  section  on  the 
xy-plane  is  x^  —  y^  =  a^. 

2.    The  integral  of  ydx  or  Vx"^  —  a^  dx  is 


Fia.  »4. 


^  xVx^  —  a2  —  1  a'^  loge  (x  +  Vx^  _  a^). 
Taking  x  for  the  upper  limit  and  a  for  the  lower,  and  putting  y  for 


Vx^  —  a^,  the  area  is  I  xy 


+  y\ 

a     I 


162  HYPERBOLIC   FUNCTIONS. 

From  P,  drop  a  perpendicular  PM  on  OA.     Then  PM 

PM 

=  a  sin  6,   0M=  a  cos  $,  and  — —  =  tan  6,  and  so  on.     From 

OM 

these  properties  of  PM  and  OiHf,  the  trigonometric  functions 
have  been  also  called  circular  functions. 

From  analogy,  certain  ratios  of  the  semi-axis,  the  abscissa, 
and  ordinate  of  a  point  on  an  equilateral  hyperbola  are 
called  hyperbolic  functions. 

They  are  defined  as  follows:  Let  u  be  such  a  quantity 
that 

Area  sector  OAQ  =  \d^u.     Then 

1.  The  ratio  of  NQ  to  OA  is  called  the  hyperbolic  sine 
of  u.  Hyperbolic  sine  u  will  be  written  sinh  u  and  will  be 
read  "  shin  ^^."     It  follows  that  QN=  a  sinh  u. 

2.  The  ratio  of  ON  to  OA  is  called  the  hyperbolic  cosine 
of   u ;    written   and   pronounced   cosh  u.      It   follows   that 

3.  The  ratio  of  sinh  u  to  cosh  u  {NQ  :  ON)  is  the  hyperbolic 
tangent  of  u  (tanh  u). 

4.  The  reciprocal  of  cosh  u  {OA :  ON)  is  the  hyperbolic 
secant  of  u  (sech  u). 

5.  The  reciprocal  of  sinh  ?*  {NQ :  OA)  is  the  hyperbolic 
cosecant  of  u  (csch  u). 

6.  The  reciprocal  of  tanh  it  {ONiNQ)  is  the  hyperbolic 
cotangent  of  u  (coth  u). 

85.   Exponential  values  of  hyperbolic  functions.  —  By  the 

property  of  the  equilateral  hyperbola, 

ON'  -  NQ"  =  OA'. 
Dividing  by  OA',  cosh^  u  —  smlci'  u  =  1. 

But  since  u^log,(2^^y 

cosh  u  -f  sinh  u  =  e". 
Hence,  by  division, 

cosh  u  —  sinh  u  =  e~". 


RELATIONS  OF  THE  FUNCTIONS.        163 

Therefore,  by  subtraction  and  addition, 

sinh?^  =  |(e«-e-«),  (83) 

cosh  u  =  ^  (e"  +  e-").  (84) 

GENERAL  RELATIONS. 
86.   Relations  of  hyperbolic  and  circular  functions. 

Now  ^  (e"  -  e- ")  =  -  i  [~  (e""  -  e" «")1 . 

Hence  by  (81)  and  (83)  :     sinh  u  =  —  i  sin  ui.  (85) 

Also,  |(e"  +  e-")  =  ^  (e""  +  e-«'*). 

Hence  by  (82)  and  (84)  :     cosh  u  =  cos  ui.  (86) 

Periodicity  of  hyperbolic  functions. 

Now  —  I  sin  ui  =  —  i  sin  (ui  —  2  mr) 

=  —  i  sin  (u  +  2  mri)  L 
By  (85),  sinh  u  =  sinh  (w  +  2  nTrt').  (87) 

Similarly,  cosh  u  =  cosh  (u-{-2  mri).  (88) 

The  addition  formulae.  —  From  (S5),  (S6),  (65),  and  (66)  : 
sinh  (u  -\-  V)  =  —  i  sin  (ui  -f-  vi) 

=  —  i  (sin  ui  cos  vi  +  cos  ui  sin  vi) 

=  sinh  u  cosh  v  -|-  cosh  u  sinh  v.  (89) 
cosh  (u-\-v)  =  cos  (wi + vi) = cos  wi  cos  vi — sin  ?i i  sin  vi 

=  cosh  w  cosh  V  +  sinh  m  sinh  v.  (90) 

Functions  of  multiples  of  u.  —  From  (89)  and  (90)  there 
is  obtained 

sinh  2u  =  2  sinh  u  cosh  u, 

cosh  2  u  =  cosh^  ^i  +  sinh^  ?/, 

sinh  3  w  =  4  sinh^  2t  +  3  sinh  w,  (91) 

cosh  3  I*  =  4  cosh^  u  —  3  cosh  u.  (92) 


164 


HYPERBOLIC   FUNCTIONS. 


87.   Longitude  of  u.  — With,  center  N  and  radius  equal  to 
NQ,  draw  an  arc  intersect-    ^  v  >  Q 

ing  the  circle  in  P.     Join 
NP  and  OP.     Since 


Fig.  95. 


(93) 
(94) 


ON'-NP'  =  OP\ 

Therefore,  the  angle  OPN 
is  a  right  angle.  Denote 
the  angle  PON  by  x. 

sinh  u  =  NQ:OA  =  NP:  0P=  tan  x, 
cosht^  =  ON:  OA  =  ON:  0P=  sec  x. 

In  (11)  put  90°  4-  X  for  x ;  then 

tan  x-\-aecx  =  tan  (45°  -f-  ^  x). 

Therefore     tan  (45°  -\-^x)=  sinh  u  +  cosh  u  =  e". 

Hence     log  tan  (45°  + 1  a;)  =  i^  log  e  =  .43429  u. 
From  which  for  any  value  of  u,  x  can  be  found,  and  thence 
in  (93)  and  (94),  sinh  u  and  cosh  u  can  be  found. 

The  angle  x  was  called  by  Gudermann  the  longitude  of  u. 
Some  later  writers,  however,  have  referred  to  it  as  the 
Gudermannian  function  of  u. 

A  table  at  the  bottom  of  page  vii  gives  values  of  u  corre- 
sponding to  values  of  x,  the  longitude  of  u,  within  limits 
which  admit  of  interpolation. 

88.  Solution  of  cubic  equations.  —  Consider  the  cubic  equa- 
tion 10^  ±  aw  =  b. 

First.     The  coefficient  of  w  is  positive. 

Let  w  =  2  sinh u  -i/ ^• 

The  given  equation  reduces  to 
sinh^  u  + 


(95) 


X  smh  u  =  —- 
Sa 


Hence  by  (91), 


sinh  3  u 


b     /3^ 
a  ^a 


(96) 


CUBIC  EQUATIONS.  165 

Denote  three  values  obtained  from  this  equation  by  3  u, 
3  w  +  2  TTi,  Su  —  2  iri. 
By  (89), 

sinh(w  ±  I  iri)=  sinh  u  cosh  f  tti  ±  cosh  u  sinh  | iri 
=  sinh  u  cos  120°  ±  i  cosh  u  sin  120°. 
Hence,  to  solve  the  equation  under  consideration,  find  u 
from  (96)  ;  then  by  (95)  the  three  roots  of  iv  are 

2  sinh u  a/^,  and  ^^  (  —  sinh  u  ±  i V3  cosh  u). 

Second.     Consider  the  equation  w^  —  aw  =  b. 
Case  I.     27  b^  is  greater  than  4  a^. 

Let  m;  =  2  cosh  w  a  Z^- 

The  given  equation  reduces  to 

cosh^  w  —  f  cosh  i^  =  TT—  \/— 
8a  ^a 

Hence  by  (92),  cosh  3u  =  ^  J?.  (97) 

Find  T^  from  (97)  ;  then  the  three  roots  are 

2  cosh  u  a/^,  and   a/-  (—  cosh  u  ±  i V3  sinh  it). 

Case  II.     27  6^  is  less  than  4  a^.  —  This  is  the  irreducible 
case  of  Tartaglia's  solution,  in  which  all  the  roots  are  real. 

Let  w  =  2  sin  x  a/-* 

The  given  equation  reduces  to 


sin'  x  —  ^smx  =  — 
8 


a  ^a 

Hence  by  (80),  sin  3  a:  =  - 1-^  ^.  (98) 

Find  X  from  (98)  ;  then  the  three  roots  of  iv  are 

2smxy[^,   2  sin  (a;  ±  120°)  ^. 


166  HYPERBOLIC  FUNCTIONS. 

Exercises.  —  1.   Find  the  value  of  i\ 

Kaise  to  the  i  power  the  equation  cos  ^  +  /  sin  ^  =  e^\ 


(cos  0  +  i  sin 

i$y  =  e-'. 

.  +  i)T. 

i'  =  e-l'^  X  e^"'' 

=  .2079e2n- 

where  n  is  any  whole  number ;  when  n  is  zero,  i*  =  .2079. 

2.  Solve  by  Horner's  Method  the  equation 

and  thus  find  sin  10°,  sin  50°,  and  sin  130°. 

3.  By  substituting  ^  for  0  in  (81),  find  sin  10°. 

18 

4.  Solve  the  following  cubic  equations : 

a?  +  2Ax  =  511  y?-   Q>x=     9  t?  -   9aj  =  10 

a^_,_54^^288  a:^-    7a;=    90  ar^-    7a;=    6 

a;3  +  72a;  =  400     .  a^-48a;  =  520  ar^-39a;  =  70 


PROBLEMS   FOR  EXAMINATION. 


[No  answers  are  given  for  these  problems.] 


RIGHT  TRIANGLES. 

154.  A  person  standing  at  a  distance  of  143  feet  from  the 
middle  of  tlie  base  of  a  tower  finds  the  altitude  of  the  tower 
to  be  43°  10'.  Find  the  height  of  the  tower,  the  eye  of  the 
observer  being  5  feet  from  the  ground. 

155.  From  the  top  of  a  lighthouse  127  feet  above  the 
level  of  the  sea,  the  angle  of  depression  of  a  boat  was  seen 
to  be  10°  20'.  Find  the  distance  of  the  boat  from  the  foot 
of  the  lighthouse. 

156.  At  what  angle  do  we  ascend  a  regular  acclivity  6.5 
miles  long,  attaining  an  altitude  at  the  summit  of  4268  feet  ? 

157.  The  horizontal  distance  between  two  columns  is  124 
feet,  and  the  straight  line  joining  the  tops  of  the  columns 
makes  an  angle  of  12°  40'  with  the  horizon.  Find  the  dis- 
tance between  the  tops. 

158.  A  canal  boat  is  drawn  along  the  middle  of  a  straight 
canal  by  two  men  hauling  upon  ropes  60  feet  and  70  feet 
long,  respectively.  Find  the  ratio  of  the  strains  upon  them, 
the  canal  being  20  feet  wide. 

159.  A  row-boat,  pulled  at  the  rate  of  5  miles  an  hour,  is 
required  to  cross  a  river  1.3  miles  wide  to  a  poi-nt  which  is 
.78  mile  lower  down  the  stream  than  the  point  of  starting. 
What  angle  must  the  direction  of  the  boat's  head  make  with 
the  bank,  supposing  the  stream  to  run  at  the  rate  of  4  miles 
an  hour  ? 

167 


168        PROBLEMS  FOR  EXAMINATION. 

160.  If  d  is  the  distance  (in  degrees)  of  the  sun  from 
the  equator,  show  that  the  altitude  of  the  sun  at  midday  is 
90°  -f-  d  —  Z,  where  I  is  the  latitude  of  the  place.  Hence, 
find  the  length  of  the  shadow  at  midday  of  a  tower  120  feet 
high,  the  latitude  of  the  tower  being  40°,  and  the  sun  being 
vertically  above  a  point  in  10°  north  latitude. 

161.  At  midday  of  the  21st  of  March,  the  length  of  the 
shadow  of  a  12-foot  pole  was  10.6  feet.  What  was  the  lati- 
tude of  the  place  ? 

162.  The  problem  of  Eratosthenes.  At  midday,  when  the 
sun  was  vertically  over  Syene,  the  shadow  of  a  vertical  rod 
at  Alexandria  was  measured,  the  shadow  being  found  to  be 
.13  times  the  length  of  the  rod.  Taking  Alexandria  to  be 
5000  stadia  due  north  of  Syene,  find  the  circumference  of 
the  earth. 

163.  The  latitude  of  Berlin  is  52°  30*.  Taking  the  earth 
to  be  a  sphere  whose  radius  =  3960  miles,  find  the  radius 
and  length  of  1°  of  the  parallel  passing  through  Berlin. 

164.  What  will  be  the  visual  angle  of  a  sphere  6  feet  in 
diameter,  if  the  eye  of  the  observer  is  placed  7  feet  from  its 
center  ? 

165.  On  a  hill  over  the  sea  is  a  60  feet  high  tower.  From 
a  boat,  the  angle  of  elevation  of  the  foot  of  the  tower  is  14°, 
and  of  the  top  of  the  tower  is  20°.  Find  the  height  of  the 
hill. 

166.  A  flagpole,  20  feet  high,  is  on  the  top  of  a  tower. 
At  a  horizontal  distance  of  420  feet  from  the  foot  of  the 
tower,  the  angle  of  elevation  of  the  highest  point  of  the 
flagpole  is  seen  to  be  15°  20'.     Find  the  height  of  the  tower. 

167.  The  lu7ie  of  Hippocrates.  A  semi-circumference  and 
a  quadrant  concave  in  the  same  direction  are  drawn  to  meet 
at  their  extremities  so  as  to  inclose  space ;  the  radius  of  the 
semi-circumference  is  10  feet ;  find  the  area  inclosed. 


RIGHT  TRIANGLES.  169 

168.  On  one  side  of  an  angle  A  =  50°,  a  distance  AB  =  10 
feet  is  taken.  From  B,  a  perpendicular  BC  is  let  fall  upon 
the  other  side  of  the  angle ;  from  C,  a  perpendicular  CD  is 
let  fall  upon  the  first  side ;  from  D  a  perpendicular  DE  is 
let  fall  upon  the  second  side ;  and  so  on,  ad  infinitum.  Find 
the  sum  of  all  the  perpendiculars. 

Put  A  instead  of  50°  and  a  instead  of  10  feet ;  then  obtain 
the  general  solution  of  this  problem.* 

169.  A  circle  has  a  radius  of  22  inches.  From  the  same 
point  in  the  circumference,  two  chords  are  drawn  of  lengths 
10  inches  and  17  inches  respectively.  What  angle  is  formed 
by  these  chords  ? 

170.  Two  towers  stand  100  feet  apart  on  the  same  hori- 
zontal plane.  An  observer  in  line  with  them,  and  150  feet 
from  the  nearer  one,  sees  the  angle  of  elevation  of  both  of 
them  to  be  the  same.  He  then  walks  directly  toward  them 
a  distance  of  130  feet,  when  he  finds  the  angle  of  elevation 
of  the  nearer  one  to  be  twice  as  great  as  that  of  the  farther 
one.     Find  the  heights  of  the  towers. 

Before  performing  the  computation  for  this  problem, 
obtain  the  general  solution  of  it. 

171.  The  top  of  a  hill  is  90  feet  higher  than  the  foot  of  a 
tower ;  from  the  top  of  the  hill  the  angle  of  depression  of 
the  top  of  the  tower  is  5°43|-'.  The  horizontal  distance 
from  the  tower  to  the  hill  is  399  feet.  Find  the  height  of 
the  tower. 

172.  I  bought  a  lot  having  the  shape  of  a  right  triangle ; 
the  hypotenuse  =  875  feet,  and  one  of  the  acute  angles  = 
37°  53'.  How  much  was  paid  for  the  land  if  the  price  was 
2.7  cents  per  square  foot  ? 

*  If  for  the  data  of  a  problem,  literal  values  are  assumed  instead  of 
numerical  values,  the  answer  to  the  problem  expressed  in  terms  of 
these  literal  values  is  called  the  general  solution  of  the  problem. 


170  PROBLEMS  FOR   EXAMINATION. 


OBLIQUE   TRIANGLES. 

173.  To  measure  the  distance  between  two  points  A  and 
B  separated  by  a  pond,  a  third  point  O  was  chosen.  AC 
was  found  to  be  545  feet,  CB  =  305  feet,  and  the  angle 
ACB  =  67°.     Find  AB. 

174.  From  a  point  on  the  bank  of  a  river,  an  observer 
goes  directly  away  from  the  river  165  yards  up  a  slope 
inclined  at  an  angle  of  12°  12'  to  the  horizontal;  at  this 
point  the  angle  of  depression  of  an  object  on  the  opposite 
bank  and  in  the  same  plane  with  the  first  position  was 
found  to  be  2°  10'.     Find  the  breadth  of  the  river. 

175.  The  range  finder.  From  a  fort  a  torpedo-boat  is 
observed  to  be  exactly  east;  eighty  seconds  afterward 
it  is  exactly  northeast,  and  eighty  seconds  after  that  it  is 
N.  22°  30'  E.  The  torpedo-boat  is  going  at  a  uniform  rate 
in  a  straight  line.  Find  its  bearing  at  the  end  of  the  next 
eighty  seconds. 

176.  A  vertical  tower  stands  on  a  plane  inclined  11°  20' 
to  the  horizontal.  Measuring  from  the  base  of  the  tower 
down  the  slope  an  oblique  distance  of  59.18  feet,  the  tower 
was  found  to  subtend  an  angle  of  38°.  Find  the  height  of 
the  tower. 

177.  Two  inaccessible  points  X  and  T  are  observed  from 
two  stations  A  and  B,  500  feet  apart  {A  and  X  on  the  left). 
The  angle  XAY=  42°,  angle  YAB  =  67°,  angle  ABX  =  39^ 
angle  X5  r=  47°.     Find  XY. 

178.  A  house  is  seen  from  two  ships  A  and  B  which  are 
300  feet  apart.  At  A,  the  angle  of  elevation  of  the  house 
is  found  to  be  28°  20'  and  the  horizontal  angle  from  it  to  the 
ship  B  =  48°  40'.  At  B,  the  horizontal  angle  from  the  house 
to  the  ship  A  is  found  to  be  20°  15'.  Find  the  height  of  the 
house  above  the  level  of  the  water. 


OBLIQUE   TRIANGLES.  171 

179.  To  find  the  length  BC  of  a  lake  running  due  north 
and  south  {C  north  of  B),  a  point  D  was  taken  530  feet 
north  of  C  and  a  point  A  was  found  due  west  of  B.  It  was 
found  that  AC  =  1970  feet  and  AD  =  2400  feet.  Find  the 
length  of  the  lake. 

180.  To  find  the  distance  between  two  inaccessible  points 
A  and  B,  an  observer  chose  a  point  C  in  the  same  plane  and 
found  angle  ACB  =  113°.  Then  he  walked  on  the  continua- 
tion of  AC  to  a  point  125  yards  from  C,  where  the  angle 
subtended  by  AB  was  found  to  be  one-half  of  angle  ACB. 
Then  he  walked  on  the  continuation  of  BC  173  yards  from 
C  and  again  found  the  angle  subtended  by  AB  to  be  one-half 
of  angle  ACB.     Find  AB. 

181.  Given  the  three  sides  of  a  triangle  to  find  the  length 
from  the  vertex  to  a  point  of  trisection  of  the  base.  Take 
the  base  (c)  =  200  feet,  a  =  205,  b  =  85,  and  find  the  length 
to  the  point  of  trisection  near  a. 

182.  How  many  square  miles  of  the  earth's  surface  can 
an  observer  see  if  stationed  at  the  top  of  Mt.  Everest,  whose 
altitude  is  29,000  feet  ?  The  radius  of  the  earth  =  3960 
miles. 

183.  A  lighthouse  60  feet  high  is  on  a  cliff  over  the  sea. 
From  the  top  of  the  lighthouse  the  angle  of  depression  of  a 
boat  was  seen  to  be  16°  20' ;  and  from  the  bottom  of  the 
lighthouse,  the  angle  of  depression  of  the  boat  was  12°  10'. 
Find  the  height  of  the  cliff  and  the  distance  of  the  ship 
from  the  foot  of  the  cliff. 

184.  Wishing  to  know  the  breadth  of  a  river  from  a  point 
X  to  a  point  A,  an  observer  measures  in  the  prolongation  of 
XA  a  distance  AB  —  500  feet ;  then  from  B,  he  measures 
at  right  angles  to  XAB  a  distance  BC  =  700  feet ;  from  the 
point  C,  the  angle  XCA  was  found  to  be  24°  20'.  Find  the 
breadth  of  the  river. 


172  PROBLEMS  FOR  EXAMINATION. 

185.  P,  Q,  and  B  are  three  known  points  in  a  straight 
line;  PQ  =  104,  QP  =  296 ;  PQ  and  QR  are  observed  to 
subtend  equal  angles  =  46°  50'  at  a  certain  point  S.  Find 
SQ. 

186.  From  the  bottom  of  a  wall  20  feet  high,  the  angle 
of  elevation  of  the  top  of  a  house  =  40°  20' ;  at  the  top  of 
the  wall  the  angle  of  elevation  of  the  same  is  19°  40'.  Find 
the  height  of  the  house. 

187.  A  person  at  a  point  A,  wishing  to  know  the  distance 
of  an  inaccessible  object  C  on  the  opposite  side  of  a  river, 
lays  off  a  base  AB  =  898.8 ;  he  then  measures  the  angle 
CAB  =  63°  20'  and  angle  CBA  =  52°  40'.     Find  AC. 

188.  Of  three  towns  the  second  is  11.6  miles  from  the 
first,  the  third  is  10.5  miles  from  the  second,  and  the  first  is 
14.3  miles  from  the  third.  Find  the  angles  of  the  triangle 
whose  vertices  are  the  three  towns. 

189.  From  the  deck  of  a  ship,  10  feet  above  the  sea,  the 
angle  of  elevation  of  the  top  of  a  cliff  is  20°  20' ;  from  the 
top  of  the  mast  &o  feet  higher,  the  angle  is  12°  30'.  Find 
the  height  of  the  cliff  from  the  sea  and  the  distance  of  the 
ship  from  the  foot  of  the  cliff. 

190.  From  the  top  of  a  tower  60  feet  high,  a  column  sub- 
tends an  angle  of  28° ;  from  the  base  of  the  tower,  the  col- 
umn subtends  an  angle  of  40°.  Find  the  height  of  the 
column. 

191.  A  balloon  is  ascending  vertically  with  a  uniform 
velocity.  When  it  is  one  mile  high,  its  angle  of  elevation  is 
32° ;  twenty  minutes  later,  the  elevation  is  54°.  How  fast 
is  the  balloon  rising  ? 

192.  At  two  points,  70  and  50  feet  from  the  foot  of  a 
tower,  a  flagstaff  on  the  top  of  the  tower  subtends  the  same 
angle,  viz.,  3°.     Find  the  length  of  the  flagstaff. 


REDUCTION  TO  FIRST  QUADRANT.  173 

FUNCTIONS  OF  ANGLES. 

Find  a  function  of  an  angle  of  the  first  quadrant  equiva- 
lent to  each  of  the  following  functions : 

193.  sin  283°.  201.  sin  324°.  209.  sec  247°  42'. 

194.  cot  196°.  202.  CSC  562°.  210.  esc  176°  28'. 

195.  cos  344°.  203.  cos  897°.  211.  cot  31 2°  53'. 

196.  sec  127°.  204.  tan  631°.  212.  sec  228°  12'. 

197.  tan  274°.  205.  tan  189°  49'.  213.  tan  142°  29'. 

198.  cos  245°.  206.  cos  128°  11'.  214.  cos  218°  47'. 

199.  CSC  144°.  207.  sin  212°  14'.  215.  sin  374°  55', 

200.  cot  133°.  208.  cot  321°  37'.  216.  esc  524°  19'. 

Reduce  each  of  the  following  expressions  to  its  simplest 
form: 

217.  a  cos  (180°  -x)-\-b  cos  (90°  +  x)  cot  x. 

218.  k  sin  X  cot  x  tan  (180°  +  x)  sec  (90°  -  x). 

219.  tan  (180°  -  x)  cot  (270°  -  x) -{-  cot  (90°  +  x)  cot  x. 

220.  cos  (90°  +  X)  sin  (360°  -  x)  -  sin  (270°  -  x)  cos  (-  x). 

221.  (a  +  by  sec  (360°  -x)-{a-  bf  esc  (90°  +  x). 

222.  sec  (90°  +  x)  CSC  (360°  -  x)  +  tan  (90°  -\-  x)  cot  x. 

223.  tan  a;  cot  a;  +  tan  (—  2/)  +  tan  (180°  -f-  y). 

224.  sin (180° -a;) COS (90° -t-ir) CSC (180°  + a;) CSC «. 

225.  (a^  +  b^)  sec  (270°  -x)-{-  (a'  -  b")  esc  (180°  +  x). 

226.  a  sin  (360°  -  x)  tan  (270°  +  x)  sec  (180°  -  x). 

227.  2  cot  2  x  COS  a;  +  2  sin  x  —  esc  x. 

228.  a?  sin  90°  +  2  a6  cos  180°  +  W  sec  0°. 

229.  m  sin  90°  -  n  cos  360°  +  {m  —  n)  cos  180°. 


174        PROBLEMS  FOR  EXAMINATION. 

Prove  the  truth  of  the  following  equations : 

___    sin  X  -\-  tan  x        •       , 

230.   — =  smictana;. 

cot  X  +  CSC  X 

The  proof  consists  in  reducing  the  equation  to  an  identity. 
Thus,  by  clearing  of  fractions,  both  sides  of  the  equation 
are  sin  a;  +  tan  x.  By  writing  the  work  in  reverse  order,  a 
direct  proof  of  the  given  equation  is  obtained. 

231.  tana:  =  2i^H^^^Hl+-^^^).. 

(1  +  cos  x)  (1  H-  CSC  x) 

__-  sec  x  cot  x  —  CSC  x  tan  x 

232.  CSC  a;  sec  a;  =  — • 


233.  tan  a;  —  cot  a;  = 


cos  X  —  sm  X 
1  —  2  cos^  X 


sm  X  cos  X 

234.  tan  X  +  cot  X  =  cot  x  sec^  x. 

235.  sin^  X  +  tan^  x  =  sec^  x  —  cos^  x. 

236.  sec  X  —  cos  X  =  sin  x  tan  x. 
_-_    sin  X  +  cos  a^  _  sec  a;  +  esc  x 

ZOt,      — ; • 

sin  X  —  cos  X     sec  x  —  esc  x 
238.   (sin  x  +  tan  x)  (cos  x  +  cot  a;)  =  (1  +  sin  x)  (1  +  cos  x). 
2  tan  X  2  cot  a; 


239.  sin  2  a;  = 

240.  tan  2  a;  = 


1  +  tan^  X     1  4-  cot^  x 

1 1 

1  —  tan  X     1  +  tan  x 

.2, 


^^,  o  sec'.'K  sec  a;  CSC  a; 

241.   sec  2  a;  = 


1  —  tan-  X     cot  X  —  tan  x 

242.  CSC  2x  =  ^  cot  a:  cos^  x  =  ^  (tan  a;  +  cot  x). 

243.  cot  2  a;  =  1^  (cot  X  —  tan  i»)  =  ^  sec  a;  esc  a;  —  tan  x. 
cot^  a;  —  1      cot  a?  —  tan  x 


244.   cos  2  a; 


cot^  a;  +  1      cot  a;  +  tan  x 


GONIOMETRIC   RELATIONS.  I75 

245.  sin  {x  +  y)  =  (tan  x  +  tan  y)  cos  x  cosy. 

246.  tan  (a: +  2/)=    .      ^^^'^-^^^'y 

sin  X  cos  a;  —  sin  y  cos  y 

247.  5ec(x  +  y)=     ^««^sec.v    , 

1  —  tan  X  tan  ?/ 

248.  sin  (x  —  y)^  tan  a;  -  tan  ?/ 

sin  (a;  +  y)      tan  a;  +  tan  y 

249.  tana;tan2/  =  *Hi^+i5M. 

cot  X  +  cot  2/ 

250.  tan(^  +  y)  +  tan(x-y)=/tan^^f.'/^   ■ 

1  —  tan^  a.-  tan^  ?/ 

251.  sin(a;+2/)-hcos(a;— ?/)=(sina;+cosa;)(sin!/4-cos2/). 

252.  tani(a;-2/)=^i2^^=-^iM  =  22^r:i^2^. 

cos  a;  4-  cos  y     sin  ?/  +  sin  x 

253.  tan  (45°  +  a;)  =  ^^^^  +  ^"^^  =  L±taii^. 

cos  X  —  sin  x*     1  —  tan  x 

254    tan  a;  ^  sin  (a;  +  y)  +  sin  (x  —  y) 
tan  y     sin  (a;  +  ?/)  —  sin  (x  —  y) 

255.  tan^  ^  x  +  2  cot  a;  tan \x  —  l  =  0. 

256.  tan- 1  a?  —  2  CSC  a;  tan  ^  a;  +  1  =  0. 

Explain  how  the  general  quadratic  equation  may  be  solved 
by  the  use  of  255  or  256. 

257.  tanla^  =  l±4^?^^=-^^^. 

1  +  sin  X  +  cos  X 

258.  sin  nx  =  2  sin  (^i  —  V)x  cos  x  —  sin  (n  —  2)x. 

259.  cos  nx  =  2  cos  (n  —  l)a;  cos  x  —  cos  (n  —  2)a;. 

260.  tanna:=  tan  (n  -  l)a^  +  tan  a;  ^ 

1  —  tan  (11  —  V)x  tan  x 


176        PROBLEMS  FOR  EXAMINATION. 

Solve  the  following  equations : 

261.  a  tan^  a?  +  6  =  c  sec  a;. 

262.  cos^  X  —  sin^  x  +  tan^  ^  =  f- 

263.  sin (20°  + a;) cos (20° -a;)  =  .2. 

264.  tan  X  +  cot  aj  =  4  cot  2  x. 

265.  cot  X  —  tan  x  =  4  cos  2  a;. 

266.  tan  2  a?  +  tan  3  a?  =  3  tan  a;. 

267.  1  —  cos  2  aj  =  3  sin  2  a;. 

268.  sin  X  +  cos  a;  =  V2. 

269.  16  (cot  a;  —  sec  x)  =  tan  x. 

270.  tan  aj  +  cot  a;  =  2  sec  2  a;. 

271.  cot  X  tan  2  a;  —  tan  a;  cot  2  a;  =  2. 

272.  sec^  X  -f-  tan  a;  =  3. 

273.  arctan(^+l)  =  3arctan(i\r-l). 


—  arc  tan =  ^-3-  ■ 

N-1  N+1      '^ 


274.   arc  tan^:= — r  —  arc  tan^r — 7  =  t-V  tt. 


ANSWERS. 


1.  c  tan  ^  =  102.7.  21.  aVcot^^  +  cot^i)  =  109. 

2.  a  cot  12°  40' =  387.1.  ^2    ___^L_^__  - 125 

3.  cot  a;  =2.4;  22°  37'.  '   Vcot^  i)  -  cot^  ^  ~ 

4.  23  sec 38°  =  29.19.  23.  b^ sm2  A=b' -  (a  -  cY; 

5.  880  sin  33°  3'  =  480.  '         N.  22°  32 '  E. 

6.  c  tan  ^  =  125.1.  24.  4ir=918(a- 6 -f  c)  ; 

7.  ctan^+d  =  18.  340,189. 

8.  c  (tan  ^  + tan  A)  =  102.  ^5.   6^  =  (a  +  c)^  -  4  7^;  377. 

9.  a(l+tan^cot^i)  =  76.5.  ^^-  1^.7,30.8. 
^Q          atan^       ^^^^  27.  4.844,8.72. 

tan  ^1- tan  ^           *  28.  14.25,8.75. 

11.   9^ =  500.  29.   154tan27°10'  +  5  =  84. 

tan^+tanA  30,  3034,52.94. 

12.  a  (tan  G  —  tan  Cj)  =  1.5.  ^^    ^go  ^q, 

13.  c(tanA-tan^)=128.4.  ^^    12.6,  54° 52'. 

c 

"•  cot^  +  cotA  =  ^^^'  33    27r^^os40_°^  2206 

ctany^  '         86400 

'  tan  ^.- tan  ^1  34.  2°  18.7',  .1615m.,  852.6  ft. 

16.   V^  =  150.  35.   tana?  =  ^  +  ^;  77°. 

aan^  _   *^^-^^ 

18.  (a  - /i)  cot  ^  =  3.1.  37.   aA/^i^  =  60. 

^  h—  a 

19.  cota;  =  ?2;  51°  41'.  2^ 

o^    ^                   .^o..,  38-  ^ CSC 33° 45' =  10. 

20.  tana;  =  n;  47° 44'.  27 

177 


178 


39.  13.5  CSC  58°  6' = 


ANSWERS. 
:  15.9. 


cot  20°  15' H- 2  cos  20°  15' 


=  21.8. 


41.   sec^ 


2.4;  A 


65°  23' ;  K=  10.64. 


42.   1.5fZ2sm60°sm2i^ 


8.798. 

43.  Area  =  two  triangles  —  sector  =  lic^  sin  60° ;  15. 

44.  7r62 cos ^  =  204.2.  45.  7r/fseci(7  =  33.49. 

46.  367rF2  =  ^^sin2(7csci(75  73.43. 

47.  rad  base  =  1.81  CSC  54°  18' ;   F=  36.65. 

48.  p^seGA  =  2p',  K=p^  Got  A  =  25.57. 

49.  tan^  =  i7r;  ^=57°  31'.   57.   2.1437rCot  25°43'=  14. 

50.  100°.    51.  50.   52.  48°  55'.  r,  90° 

106.27  7r6 


53. 


6.185. 


180  a 
54.  21.  55.   163°  34'. 

56.   .3125  d^csc^  36°  =  12.15. 


58. 


J  a  CSC 


a  cos 


2n  +  l 
180° 


2n  +  l 
59.   143.1  feet. 
60.   27  r"  tan  6°  40'  sin26°  40'  =  3.6. 


=  8.988. 
7.208. 


„    cyTYij^       360°.       180°     ^^^^ 

61.   2—  A^csc tan =  116.5. 

n  n  TYi 

62.  1 7^  (16.96  TT  +  sin  70°  32 '  -  sin  40°)  =  .4656. 

63.  441  cos  46°  tan  20°  =  157.     67.  c  sin  (A^-A)  sec  A  sec  A-,. 

64.  acsc(^i— J.)cos^sin^i.   68.  ccsc  (J.+J.1)  sin^sin^i. 

65.  acsc(^-|-^i)cos^cos^i.   69.  c  esc  (^— ^j)  cos^  sin^i. 

66.  a  sin  (C-  d)  sec  C  sec  Oj.     70.  Apply  (19) ;  890. 

71.  Apply  (20)  ;  741.2. 

72.  18  sec  87°  30'  cos  37°  30'  sin  40°  =  210. 

73.  50  CSC  5°  55'  cos  31°  12'  sin  25°  17'  =  177.1. 
50  esc  5°  55'  cos  25°  17'  =  438.4. 

74.  Apply  (24)  ;  N.  87°  49'  E. 
Apply  (19)  ;  12  miles  an  hour. 


OBLIQUE   TRIANGLES.  I79 

75.  528  CSC  37°  sin  42°  sec  29°  sin  8°  =  114.5. 

528  CSC  37°  sin  5°  =  301. 

76.  c  CSC  C  sin  ^  sin  J5  —  50  =  240. 

77.  775.3  CSC  27°  35'  sin  126°  43'  sin  32°  57'  =  730. 

78.  200  sec2 26°  37' CSC  10°  8' sin  10°  22' sin  20°  30' =  89.63. 

79.  180  CSC  17°  15'  sin  33°  45'  sec  40°  sin  11°  =  84. 

80.  Apply  (27) ;  18.71  -|-  16.2  =  34.91. 

81.  Apply  (24)  ;  cot  a;  =  1.76  esc  30°  -  cot  30°. 
29°  15'  with  the  front  of  the  train. 

82.  6  sin  45°  CSC  10°  =  24.44.      85.  3.5. 

83.  N.  70°  44'  E. ;  9.983.  86.   Apply  (30)  ;  229,000. 

84.  10.32;  78°  14'.  87.    Apply  (29) ;  52. 

88.  Apply  (26);  h  ^xGoiA,l  =  xcotD,  a=  XQotB\  60. 

89.  Apply  (19)  ;  20.51 ;  23.63. 

90.  Apply  (16),  (19),  and  (20)  ; 

cos^  =  -^^+i-  =  -^±i-;  c  =  5. 

2(c  +  l)      2(c-l)' 

91.  cco^\{B  +  A)  ^\n\{B  -  A)  =  51. 

92.  Snell's;  1034,1465,2047. 

93.  300  CSC  68°  14'  sin  120°  26'  csc  17°  48'  sin  70°  =  856.2. 

94.  100  sin  88°  10' csc  8°  20' =  676.      95.   Apply  (25)  ;  5.715. 
96.  Apply  (25) ;  331.  97.   Apply  (28)  ;  1054. 

98.  Find  AB  and  BD  and  apply  (20)  ;  287.2. 

99.  Find  AB,  then  BC  =  .3537. 

100.  sin  ^Z)5  =  tan  56°  26' cot  65°  23' sin  56°  33';  114. 

101.  Apply  (19)  ;  410. 

102.  Same  as  Problem  88 ;  100. 

103.  Snell's  ;  229,  61,  109. 

104.  By(30)cosM=.85;  ^=22°  47'. 

By  (28)  cot  aj=1.125  csc 45°  34'-cot45°  34';  a;=59°  15'. 
By  (19)  ^=9  csc  22°  47'  sin  82°  2'  cos  59°  15'=11.77. 


180  ANSWERS. 

105.  I  (12  CSC  A)  (16  CSC  A)  sin  ^  =  96  esc  ^  =  100. 

106.  Divide  into  a  triangle  and  a  parallelogram ;  46.82. 

107.  i a ^  +  «'±V3^  =  9 sec.  or  1  min.  57  see. 

v^  +  ^t7^  —  i^it; 

108.  70  sin  1  (^1  -  A2)  sin  i  (J5i  +  ^2)  X 
esc  1  (A  +  ^2  +  ^1  +  -S2)  =  18.34. 

109.  The  pond  is  the  incircle  of  the  triangle  formed  by 
the  three  tangents  ;  36.58. 

110.  The  pond  is  the  excircle  of  the  triangle  formed  by 
the  three  tangents ;  768.3. 

111.  The  hill  is  at  the  center  of  the  circumcircle. 
60.8  CSC  15°  25'  tan  52°  40'  =  300. 

112.  r  tan  ^5  tan  1(7=1.589.      113.   r  tan^  (45 -i^) =1.718. 

114.  4r  sin  (45°  -  ^A)  sin  (45°  -\B)  sin  (45°-J  C)=2.925. 

115.  3107.  116.   5057.  117.   2713. 
118.   2410  geographical  miles.  119.   7  hours. 

120.  lat.  37°  36' ;  long.  68°  38'. 

121.  cosZ  =  -secZsind     238° 42';  270°;  301°  18';  180°. 
cos  P  =  —  tan  I  tan  d.     length  =  2  P  -r- 15. 

14  hr.  50  min.  48  sec. ;  12  hr. ;  9  hr.  9  min.  12  sec. ;  24  hr. 

122.  sec  P  =  tan  I  cot  d.    3  hr.  55  min.  28  sec. ;  6  ;  12  p.m. 

123.  sin  a  =  sin  Z  sin  d.  14°  49 '.     tan  Z=secl  cot  d.  251°  37'. 

124.  tan  Z  =  tan  d  sec  31°  30';  14°  46'. 

125.  Let  R  be  first  position,  S  the  secon'd. 

i?^  =  49°40',  PP*S  =  84°2',  ZP^  =  50°31';   lat.  =  47°  55'. 

126.  P  =  32°  30',  ZS  =  45°  40',  PS  =  116°  6' ;  lat.  =  6°  56'. 

127.  40°  6'.  128.   .336  PI  129.   2.443  P2. 

130.  fl9K±Z^M±= -1.121.       131.  -^ ^-^^ =.216. 

6  log  a— e  log  d  6  loga+dlog  c 

132.   Eoots  of  x^  \ogd  —  x  log  5  =  log  a  —  log  c.     Imaginary. 


GONIOMETRY.  181 

133.  ^  ^  log  ?>  (/log  ^  -  ^i  log  ^)  ^  _  21.93. 

log  a  log  e  —  log  h  log  c 

134.  „  =  iri2g*  +  M  =  1.108. 

2  |_log  a     log  6J 

135.  Eoots  of  x^  log  a  +  a;  log  6  +  log  c  =  0.     Imaginary. 

136.  loglogc-loglogg^  g3o^ 

log  6 

,^„    ,  .6  sin  20°  -  sin  10°    ^o  KAf 

137.  tan  a;  = -— — -;  2°  56'. 

cos  10° -.6  cos  20°' 

138.  Reduce  to  sines  and  cosines  and  apply  (69), 

sin  {2x-  10°)  =  -  5  sin  10° ;  125°  7'. 

139.  Apply  the  value  of  tan  (x  ±  y)  and  reduce 

cot2a;  =  l+2tan2  10°;  44° 8'. 

140.  Apply  (65)  and  (75) ;  cot  a;=- (2  + cot  10°);  172°  34'. 

141.  cos  (2x-  10°)  =  1.2  -  cos  10° ;  43°  47'. 

142.  By  (13)  ;  32  cos^  a;  =  16  +  16  cos  2  x. 
16cos22a;  +  16cos2a;  +  3  =  0;  a;  =  52°14'. 

143.  Apply  (69);  29°  33'.     144.  4cosa;=  V17-1;  38°  40'. 

145.  sin  (2  a; -70°)  =  .6  CSC  70°;  105°  10'. 

146.  tan x= .4142 ;  22°  30'.     147.   tan  x  =  tan  a  tan  h  tan  c. 

148.  160  tan  ^  =  70  tan  {A  +  21°)  =  168. 

149.  a  tan  ^  =  (a  —  b)  tan  3  ^  =  59.2. 

150.  97°  37'  with  the  road;  11.44  miles. 

151.  360  cot  A  =  360  cot  (^  -  2°  7 ')  =  4994. 

152.  ir'iO-  sin  0)  =  13.714  and  4.252 ;  9.462. 
7rr2  -  (13.714  +  4.252)  =  1239. 

153.  Two  segments,;  0,  =  153°  44',  6^  =  54°  33';  .1  A. 


182  FORMULAE   FOR   RIGHT  TRIANGLES. 


tan  A 


COS  A 


PLANE   RIGHT  TRIANGLES. 

Any  side  =  adjacent  side  times  adjacent  function. 

Any  function  of  A  =  adjacent  side  divided  by  the  next  side. 

Any  function  of  C  =  the  corresponding  cofunction  of  A. 


SPHERICAL   RIGHT   TRIANGLES. 


Any  function  is  equal  to 
the  product  of  its  adjacent 
functions. 

Any  function  is  equal  to 

the  product  of  an  adjacent 

function  and  the  reciprocal 

of  the  next  function.  ^'"^ 

tan  A~-^ 


tape 


—CSC  A— sin 


fan  a — cos  G- 


A   TABLE   OF 

FOUR-PLACE    MANTISSA    OF    LOGARITHMS 
OF   NUMBERS 

FROM  1  TO  1000 


A  TABLE   OF 

THE   LOGARITHMS   OF  THE    TRIGONOMETRIC 
FUNCTIONS  OF  ANGLES 

FROM  0°  TO  90°  FOR   EVERY   10' 


A  TABLE  OF  FORMULA 


A  TABLE  OF  VALUES  OF  CONSTANTS 


viu 


LOGARITHMS  OF  NUMBERS. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8    9 

10 

.0000 

.0043 

.0086 

.0128 

.0170 

.0212 

.0253 

.0294 

.0334  .0374 

11 

414 

453 

492 

531 

569 

607 

645 

682 

719  755 

12 

792 

828 

864 

899 

934 

969 

.1004 

.1038 

.1072  .1106 

13 

.1139 

.1173 

.1206 

.1239 

.1271 

.1303 

335 

367 

399  430 

14 

461 

492 

523 

553 

584 

614 

644 

673 

703   732 

15 

.1761 

.1790 

.1818 

.1847 

.1875 

.1903 

.1931 

.1959 

.1987  .2014 

16 

.2041 

2068 

.2095 

.2122 

.2148 

.2175 

.2201 

.2227 

.2253  279 

17 

304 

330 

355 

380 

405 

430 

455 

480 

504  529 

18 

553 

577 

601 

625 

648 

672 

695 

718 

742  765 

19 

788 

810 

833 

856 

878 

900 

923 

945 

967  989 

20 

.3010 

.3032 

.3054 

.3075 

.3096 

.3118 

.3139 

.3160 

.3181  .3201 

21 

222 

243 

263 

284 

304 

324 

345 

365 

385  404 

22 

424 

444 

464 

483 

502 

522 

541 

560 

579  598 

23 

617 

636 

655 

674 

692 

711 

729 

747 

766  784 

24 

802 

820 

838 

856 

874 

892 

909 

927 

945  962 

25 

.3979 

.3997 

.4014 

.4031 

.4048 

.4065 

.4082 

.4099 

.4116  .4133 

26 

.4150 

.4166 

183 

200 

216 

232 

249 

265 

281  298 

27 

314 

330 

346 

362 

378 

393 

409 

425 

440  456 

28 

472 

487 

502 

518 

533 

548 

564 

579 

594  609 

29 

624 

639 

654 

669 

683 

698 

713 

728 

742  757 

30 

.4771 

.4786 

.4800 

.4814 

.4829 

.4843 

.4857 

.4871 

.4886  .4900 

31 

914 

928 

942 

955 

969 

983 

997 

.5011 

.5024  .5038 

32 

.5051 

.5065 

.5079 

.5092 

.5105 

.5119 

.5132 

145 

159  172 

33 

185 

198 

211 

224 

237 

250 

263 

276 

289  302 

34 

315 

328 

340 

353 

366 

378 

391 

403 

416  428 

35 

.5441 

.5453 

.5465 

.5478 

.5490 

.5502 

.5514 

.5527 

.5539  .5551 

36 

563 

575 

587 

599 

611 

623 

635 

647 

658  670 

37 

682 

694 

705 

717 

729 

740 

752 

763 

775  786 

38 

798 

809 

821 

832 

843 

855 

866 

877 

888  899 

39 

911 

922 

933 

944 

955 

966 

977 

988 

999  .6010 

40 

.6021 

.6031 

.6042 

.6053 

.6064 

.6075 

.6085 

.6096 

.6107  .6117 

41 

128 

138 

149 

160 

170 

180 

191 

201 

212  222 

42 

232 

243 

253 

263 

274 

284 

294 

304 

314  325 

43 

335 

345 

355 

365 

375 

385 

395 

405 

415  425 

44 

435 

444 

454 

464 

474 

484 

493 

503 

513  522 

45 

.6532 

.6542 

.6551 

.6561 

.6571 

.6580 

.6590 

.6599 

.6609  .6618 

46 

628 

637 

646 

656 

665 

675 

684 

693 

702  712 

47 

721 

730 

739 

749 

758 

767 

776 

785 

794  803 

48 

812 

821 

830 

839 

848 

857 

866 

875 

884  893 

49 

902 

911 

920 

928 

937 

946 

955 

964 

972  981 

50 

.6990 

.6998 

.7007 

.7016 

.7024 

.7033 

.7042 

.7050 

.7059  .7067 

51 

.7076 

.7084 

093 

101 

110 

118 

126 

135 

143   152 

52 

160 

168 

177 

185 

193 

202 

210 

218 

226  235 

53 

243 

251 

259 

267 

275 

284 

2.2 

300 

308  316 

54 

324 

332 

340 

348 

356 

364 

372 

380 

388  396 

LOGARITHMS   OF  NUMBERS. 


IX 


N 
55 

0 

1 

2    3    4 

5    6 

7    8    9 

.7404 

.7412 

.7419  .7427  .7435 

.7443  .7451 

.7459  .7466  .7474 

56 

482 

490 

497  505  513 

520  528 

536  543  551 

57 

559 

566 

574  582  589 

597  604 

612  619  627 

58 

634 

642 

649  657  664 

672  679 

686  694  701 

59 

709 

716 

723  731   738 

745  752 

760  767  774 

60 

.7782 

.7789 

.7796  .7803  .7810 

.7818  .7825 

.7832  .7839  .7846 

61 

853 

860 

868  875  882 

889  896 

903  910  9i7 

62 

924 

931 

938  945  952 

959  966 

973  980  987 

63 

993 

.8000 

.8007  .8014  .8021 

.8028  .8035 

.8041  .8048  .8055 

64 

.8062 

069 

075  082  089 

096  102 

109  116  122 

65 

.8129 

.8136 

.8142  .8149  .8156 

.8162  .8169 

.8176  .8182  .8189 

66 

195 

202 

209  215  222 

228  235 

241  248  254 

67 

261 

267 

274  280  287 

293  299 

306  312  319 

68 

325 

331 

338  344  351 

357  363 

370  376  382 

69 

388 

395 

401  407  414 

420  426 

432  439  445 

70 

.8451 

.8457 

.8463  .8470  .8476 

.8482  .8488 

.8494  .8500  .8506 

71 

513 

519 

525  531  537 

543   549 

555  561   567 

72 

573 

579 

585  591  597 

603  609 

615  621  627 

73 

633 

639 

645  651  657 

663  669 

675  681  686 

74 

692 

698 

704  710  716 

722  727 

733  739  745 

75 

.8751 

.8756 

.8762  .8768  .8774 

.8779  .8785 

.8791  .8797  .8802 

76 

808 

814 

820  825  831 

.  837  842 

848  854  859 

77 

865 

871 

876  882  887 

893  899 

904  910  915 

78 

921 

927 

932  938  943 

949  954 

960  965  971 

79 

976 

982 

987  993  998 

.9004  .9009 

.9015  .9020  .9025 

80 

.9031 

.9036 

.9042  .9047  .9053 

.9058  .9063 

.9069  .9074  .9079 

81 

085 

090 

096  101   106 

112  117 

122   128  133 

82 

138 

143 

149  154  159 

165   170 

175  180  186 

83 

191 

196 

201  206  212 

217  222 

227  232  238 

84 

243 

248 

253  258  263 

269  274 

279  284  289 

85 

.9294 

.9299 

.9304  .9309  .9315 

.9320  .9325 

.9330  .9335  .9340 

86 

345 

350 

355  360  365 

370  375 

380  385  390 

87 

395 

400 

405  410  415 

420  425 

430  435  440 

88 

445 

450 

455  460  465 

469  474 

479  484  489 

89 

494 

499 

504  509  513 

518  523 

528  533  538 

90 

.9542 

.9547 

.9552  .9557  .9562 

.9566  .9571 

.9576  .9581  .9586 

91 

590 

595 

600  605  609 

614  619 

624  628  633 

92 

638 

643 

647  652  657 

661  666 

671  675  680 

93 

685 

689 

694  699  703 

708  713 

717  722  727 

94 

731 

736 

741  745  750 

754  759 

763  768  773 

95 

.9777 

.9782 

.9786  .9791  .9795 

.9800  .9805 

.9809  .9814  .9818 

96 

823 

827 

832  836  841 

845  850 

854  859  863 

97 

868 

872 

877,,  881  886 
92  r  926  930 

890  894 

899  903  908 

98 

912 

917 

934  939 

943  948  952 

99 

956 

961 

965  969  974 

978  983 

987  991  996 

LOGAKITHMS   OF   FUNCTIONS. 


X 

L  sivLX 

/.tanjr 

log  sec  X 

log  CSC  X 

log  cot  X 

L  cosjr 

0" 

Neg.  Inf. 

Neg.  Inf 

0  000  0 

Infinite 

Infinite  10.000  0 

90= 

10' 

7.  4637 

7.  4637 

0 

2.  5363 

2.  5363 

0 

50' 

20' 

7648 

7648 

0 

2352 

2352 

0 

40' 

^0' 

9408 

9409 

0 

0592 

0591 

0 

30' 

40' 

8.  0658 

8.  0658 

0 

1.  9342 

1.  9342 

0 

20' 

50' 

1627 

1627 

0 

8373 

8373 

0 

10' 

1= 

8.  2419 

8.  2419 

0.000  1 

1.  7581 

1.  7581 

9.999  9 

89° 

10' 

3088 

3089 

1 

6912 

6911 

9 

50' 

20' 

3668 

3669 

1 

6332 

6331 

9 

40' 

30' 

4179 

4181 

1 

5821 

5819 

9 

30' 

40' 

4637 

4638 

2 

5363 

5362 

8 

20' 

50' 

5050 

5053 

2 

4950 

4947 

8 

10' 

2° 

8.  5428 

8.  5431 

0.000  3 

1.  4572 

1.  4569 

9.999  7 

88° 

10' 

5776 

5779 

3 

4224 

4221 

7 

50' 

20' 

6097 

6101 

4 

3903 

3899 

6 

40' 

30' 

6397 

6401 

4 

3603 

3599 

6 

30' 

40' 

6677 

6682 

5 

3323 

3318 

5 

20' 

50' 

6940 

6945 

5 

3060 

3055 

5 

10' 

3° 

8.  7188 

8.  7194 

0.000  6 

1.  2812 

1.  2806 

9.999  4 

87° 

10' 

7423 

7429 

7 

2577 

2571 

3 

50' 

20' 

7645 

7652 

7 

2355 

2348 

3 

40' 

30' 

7857 

7865 

8 

2143 

2135 

2 

30' 

40' 

8059 

8067 

9 

1941 

1933 

1 

20' 

50' 

8251 

8261 

0.001  0 

1749 

1739 

0 

10' 

40 

8.  8436 

8.  8446 

0.001  1 

1.  1564 

1.  1554 

9.998  9 

86° 

10' 

8613 

8624 

1 

1387 

1376 

9 

50' 

20' 

8783 

8795 

2 

1217 

1205 

8 

40' 

30' 

8946 

8960 

3 

1054 

1040 

7 

30' 

40' 

8.  9104 

8.  9118 

4 

1.  0896 

1.  0882 

6 

20' 

50' 

9256 

9272 

5 

0744 

0728 

5 

10' 

5^-^ 

8.  9403 

8.  9420 

0.001  7 

1.  0597 

1.  0580 

9.998  3 

85° 

10' 

9545 

9563 

8 

0455 

0437 

2 

50' 

20' 

9682 

9701 

9 

0318 

0299 

1 

40' 

30' 

9816 

9836 

0.002  0 

0184 

0164 

0 

30' 

40' 

9945 

9966 

1 

0055 

0034 

9.997  9 

20' 

50' 

9.  0070 

9.  0093 

3 

0.  9930 

0.  9907 

7 

10' 

6^ 

9.  0192 

9.  0216 

0.002  4 

0.  9808 

0.  9784 

9.997  6 

84° 

10' 

0311 

0336 

5 

9689 

9664 

5 

50' 

20' 

0426 

0453 

7 

9574 

9547 

3 

40' 

30' 

0539 

0567 

8 

9461 

9433 

2 

30' 

40' 

0648 

0678 

9 

9352 

9322 

1 

20' 

50' 

0755 

0786 

0.003  1 

9245 

9214 

9996  9 

10' 

T 

0859 

9.  0891 

0.003  2 

9141 

0  9109 

9.996  8 

83° 

L  cos/ 

Lcoty 

log  CSC/ 

log  sec/  log  tan/ 

L  sin/ 

/ 

LOGARITHMS  OF  FUNCTIONS. 


XI 


X 

L  sinjr 

L  tan  AT 

log  sec  X 

log  CSC  X 

Z.cot;r 

L  cosjf 

T 

9.0  859 

9.0  891 

0.00  32 

0.9  141 

0.9  109 

9.99  68 

83° 

10' 

961 

995 

34 

039 

005 

66 

50' 

20' 

9.1060 

9.1096 

36 

0.8  940 

0.8  904 

64 

40' 

30' 

157 

194 

37 

843 

806 

63 

30' 

40' 

252 

291 

39 

748 

709 

61 

20' 

50' 

345 

385 

0.00  41 

665 

615 

59 

10' 

8^ 

9.1  436 

9.1  478 

0.00  42 

0.8  564 

0.8  522 

9.99  58 

82° 

10' 

525 

569 

44 

475 

431 

56 

50' 

20' 

612 

658 

46 

388 

342 

54 

40' 

30' 

697 

745 

48 

303 

255 

52 

30' 

40' 

781 

831 

50 

219 

169 

50 

20' 

50' 

863 

915 

52 

137 

085 

48 

10' 

9^ 

9.1943 

9.1997 

0.00  54 

0.8  057 

0.8  003 

9.99  46 

sr 

10' 

9.2  022 

9.2  078 

56 

0.7  978 

0.7  922 

44 

50' 

20' 

100 

158 

58 

900 

842 

42 

40' 

30' 

176 

236 

60 

824 

764 

40 

30' 

40' 

251 

313 

62 

749 

687 

38 

20' 

50' 

324 

389 

64 

676 

611 

36 

10' 

10^ 

9.2  397 

9.2  463 

0.00  66 

0.7  603 

0.7  537 

9.99  34 

80° 

10' 

468 

536 

69 

532 

464 

31 

50' 

20' 

538 

609 

71 

462 

391 

29 

40' 

30' 

606 

680 

73 

394 

320 

27 

30' 

40' 

674 

750 

76 

326 

250 

24 

20' 

50' 

740 

819 

78 

260 

181 

22 

10' 

ir 

9.2  806 

9.2  887 

0.00  81 

0.7  194 

0.7  113 

9.99  19 

79° 

10' 

870 

953 

83 

130 

047 

17 

50' 

20' 

934 

9.3  020 

86 

066 

0.6  980 

14 

40' 

30' 

997 

085 

88 

003 

915 

12 

30' 

40' 

9.3  058 

149 

91 

0.6  942 

851 

09 

20' 

50' 

119 

212 

93 

881 

788 

07 

10' 

12^ 

9.3  179 

9.3  275 

0.00  96 

0.6  821 

0.6  725 

9.99  04 

78^ 

]0' 

238 

336 

99 

762 

664 

01 

50' 

20' 

296 

397 

0.01  01 

704 

603 

9.98  99 

40' 

30' 

353 

458 

04 

647 

542 

96 

30' 

40' 

410 

517 

07 

590 

483 

93 

20' 

50' 

466 

576 

10 

534 

424 

90 

10' 

13° 

9.3  521 

9.3  634 

0.01  13 

0.6  479 

0.6  366 

9.98  87 

77° 

10' 

575 

691 

16 

425 

309 

84 

50' 

20' 

629 

748 

19 

371 

252 

81 

40' 

30' 

682 

804 

22 

318 

196 

78 

30' 

40' 

734 

859 

2"5 

266 

141 

75 

20' 

50' 

786 

914 

28 

214 

086 

72 

10' 

140 

9.3  837 

9.3  968 

0.01  31 

0.6  163 

0.6  032 

9.98  69 

76 

L  cos/ 

L  cot/ 

log  esc/ 

log  sec/ 

log  tan/ 

L  sin/ 

/ 

Xll 


LOGARITHMS   OF  FUNCTIONS. 


X 

L  sinjr 

Ltanx 

log  sec  X 

log  CSC  X 

log  cot  X 

Z.cosjr 

14° 

9.3  837 

9.3  968 

0.01  31 

0.6  163 

0.6  032 

9.98  69 

76° 

10' 

887 

9.4  021 

34 

113 

0.5  979 

66 

50' 

20' 

937 

074 

37 

063 

926 

63 

40' 

30' 

986 

127 

41 

014 

873 

59 

30' 

40' 

9.4  035 

178 

44 

0.5  965 

822 

56 

20' 

50' 

083 

230 

47 

917 

770 

53 

10' 

15" 

9.4  130 

9.4  281 

0.01  51 

0.5  870 

0.5  719 

9.98  49 

75° 

10' 

177 

331 

54 

823 

669 

46 

50' 

20' 

223 

381 

57 

777 

619 

43 

40' 

30' 

269 

430 

61 

731 

570 

39 

30' 

40' 

314 

479 

64 

686 

521 

36 

20' 

50' 

359 

527 

68 

641 

473 

32 

10' 

16° 

9.4  403 

9.4  575 

0.01  72 

0.5  597 

0.5  425 

9.98  28 

74° 

10' 

447 

622 

75 

553 

378 

25 

50' 

20' 

491 

669 

79 

509 

331 

21 

40' 

30' 

533 

716 

83 

467 

284 

17 

30' 

40' 

576 

762 

86 

424 

238 

14 

20' 

50' 

618 

808 

90 

382 

192 

10 

10' 

17° 

9.4  659 

9.4  853 

0.01  94 

0.5  341 

0.5  147 

9.98  06 

73° 

10' 

700 

898 

98 

300 

102 

02 

50' 

20' 

741 

943 

0.02  02 

259 

057 

9.97  98 

40' 

30' 

781 

987 

06 

219 

013 

94 

30' 

40' 

821 

9.5  031 

10 

179 

0.4  969 

90 

20' 

50' 

861 

075 

14 

139 

925 

86 

10' 

18° 

9.4  900 

9.5  118 

0.02  18 

0.5  100 

0.4  882 

9.97  82 

72^ 

10' 

939 

161 

22 

061 

839 

78 

50' 

20' 

977 

203 

26 

023 

797 

74 

40' 

30' 

9.5  015 

245 

30 

0.4  985 

755 

70 

30' 

40' 

052 

287 

35 

948 

713 

65 

20' 

50' 

090 

329 

39 

910 

671 

61 

10' 

19° 

9.5  126 

9.5  370 

0.02  43 

0.4  874 

0.4  630 

9.97  57 

71° 

10' 

163 

411 

48 

837 

589 

52 

50' 

20' 

199 

451 

52 

801 

549 

48 

40' 

30' 

235 

491 

57 

765 

509 

43 

30' 

40' 

270 

531 

61 

730 

469 

39 

20' 

50' 

306 

571 

66 

694 

429 

34 

10' 

20° 

9.5  341 

9.5  611 

0.02  70 

0.4  659 

0.4  389 

9.97  30 

70° 

10' 

375 

650 

75 

625 

350 

25 

50' 

20' 

409 

689 

79 

591 

311 

21 

40' 

30' 

443 

727 

84 

557 

273 

16 

30' 

40' 

477 

766 

89 

523 

234 

11 

20' 

50' 

510 

804 

94 

490 

196 

06 

10' 

21^ 

9.5  543 

9.5  842 

0.02  98 

0.4  457 

0.4  158 

9.97  02 

69° 

L  cos/ 

L  cot/  log  CSC/ 

log  sec/  log  tan/ 

L  sin/ 

/ 

LOGARITHMS   OF   FUNCTIONS. 


Xlll 


X 

Z-sinx 

Z.tanjr 

log  sec  X 

log  CSC  X 

log  cot  X 

Z.COSX 

21° 

9.5  543 

9.5  842 

0.02  98 

0.4  457 

0.4  158 

9.97  02 

69° 

10' 

576 

879 

0.03  03 

424 

121 

9.96  97 

50' 

20' 

609 

917 

08 

391 

083 

92 

40' 

30' 

641 

954 

13 

359 

046 

87 

30' 

40' 

673 

991 

18 

327 

009 

82 

20' 

50' 

704 

9.6  028 

23 

296 

0.3  972 

77 

10' 

22° 

9.5  736 

9.6  064 

0.03  28 

0.4  264 

0.3  936 

9.96  72 

68° 

10' 

767 

100 

33 

233 

900 

67 

50' 

20' 

798 

136 

39 

202 

864 

61 

40' 

30' 

828 

172 

44 

172 

828 

56 

30' 

40' 

859 

208 

49 

141 

792 

51 

20' 

50' 

889 

243 

54 

111 

757 

46 

10' 

23° 

9.5  919 

9.6  279 

0.03  60 

0.4  081 

0.3  721 

9.96  40 

67° 

10' 

948 

314 

65 

052 

686 

35 

50' 

20' 

978 

348 

71 

022 

652 

29 

40' 

30' 

9.6  007 

383 

76 

0.3  993 

617 

24 

30' 

40' 

036 

417 

82 

964 

583 

18 

20' 

50' 

065 

452 

87 

935 

548 

13 

10' 

24° 

9.6  093 

9.6  486 

003  93 

0.3  907 

0.3  514 

9.96  07 

66° 

10' 

121 

520 

98 

879 

480 

02 

50' 

20' 

149 

553 

0.04  04 

851 

447 

9.95  96 

40' 

30' 

177 

587 

10 

823 

413 

90 

30' 

40' 

205 

620 

16 

795 

380 

84 

20' 

50' 

232 

654 

21 

768 

346 

79 

10' 

25° 

9.6  259 

9.6  687 

0.04  27 

0.3  741 

0.3  313 

9.95  73 

65° 

10' 

286 

720 

2>-^ 

714 

280 

67 

50' 

20' 

313 

752 

39 

687 

248 

61 

40' 

30' 

340 

785 

45 

660 

215 

55 

30' 

40' 

366 

817 

51 

634 

183 

49 

20' 

50' 

392 

850 

57 

608 

150 

43 

10' 

26° 

9.6  418 

9.6  882 

0.04  63 

0.3  582 

0.3  118 

9.95  37 

64° 

10' 

444 

914 

70 

556 

086 

30 

50' 

20' 

470 

946 

76 

530 

054 

24 

40' 

30' 

495 

977 

82 

505 

023 

18 

30' 

40' 

521 

9.7  009 

88 

479 

0.2  991 

12 

20' 

50' 

546 

040 

95 

454 

960 

05 

10' 

27° 

9.6  570 

9.7  072 

0.05  01 

0.3  430 

0.2  928 

9.94  99 

63° 

10' 

595 

103 

08 

405 

897 

92 

50' 

20' 

620 

134 

14 

380 

866 

86 

40' 

30' 

644 

165 

21 

356 

835 

79 

30' 

40' 

668 

196 

27 

332 

804 

73 

20' 

50' 

692 

226 

34 

308 

774 

66 

10' 

28° 

9.6  716 

9.7  257 

0.05  41 

0.3  284 

0.2  743 

9.94  59 

62° 

L  cos/ 

/.cot/ 

log  CSC/ 

log  sec/  log  tan/ 

/.sin/ 

/ 

XIV 


LOGARITHMS  OF  FUNCTIONS. 


X 

I  sin  A- 

L  tan  X   log  sec  x 

log  CSC  x 

log  cot  X 

LCOBX 

28° 

9.6  716 

9.1  2S1    0.05  41 

0.3  284 

0.2  743 

9.94  59 

62° 

10' 

740 

287     47 

260 

713 

53 

50' 

20' 

763 

317     54 

237 

683 

46 

40' 

30' 

787 

348     61 

213 

652 

39 

30' 

40' 

810 

378     68 

190 

622 

32 

20' 

50' 

833 

408     75 

167 

592 

25 

10' 

29° 

9.6  856 

9.7  438  0.05  82 

0.3  144 

0.2  562 

9.94  18 

61° 

10' 

878 

467     89 

122 

533 

11 

50' 

20' 

901 

497     96 

099 

503 

04 

40' 

30' 

923 

526  0.06  03 

077 

474  9.93  97 

30' 

40' 

946 

556     10 

054 

444 

90 

20' 

50' 

968 

585     17 

032 

415 

83 

10' 

30° 

9.6  990 

9.7  614  0.06  25 

0.3  010 

0.2  386 

9.93  75 

60° 

10' 

9.7  012 

644     32 

0.2  988 

356 

68 

50' 

20' 

033 

673     39 

967 

327 

61 

40' 

30' 

055 

701     47 

945 

399 

53 

30' 

40' 

076 

730     54 

924 

270 

46 

20' 

50' 

097 

759     62 

903 

241 

38 

10' 

31° 

9.7  118 

9.7  788  0.06  69 

0.2  882 

0.2  212 

9.93  31 

59° 

10' 

139 

816     77 

861 

184 

23 

50' 

20' 

160 

845     85 

840 

155 

15 

40' 

30' 

181 

873     92 

819 

127 

08 

30' 

40' 

201 

902  0.07  00 

799 

098 

00 

20' 

50' 

222 

930     08 

778 

070 

9.92  92 

10' 

32° 

9.7  242 

9.7  958  0.07  16 

0.2  758 

0.2  042 

9.92  84 

58° 

10' 

262 

986     24 

738 

014 

76 

50' 

20' 

282 

9.8  014     32 

718 

0.1  986 

68 

40' 

30' 

302 

042     40 

698 

958 

60 

30' 

40' 

322 

070     48 

678 

930 

52 

20' 

50' 

342 

097     56 

658 

903 

44 

10' 

33° 

9.7  361 

9.8  125  0.07  64 

0.2  639 

0.1  875 

9.92  36 

57° 

10' 

380 

153     72 

620 

847 

28 

50' 

20' 

400 

180     81 

600 

820 

19 

40' 

30' 

419 

208     89 

581 

792 

11 

30' 

40' 

438 

235     97 

562 

765 

03 

20' 

50' 

457 

263  0.08  06 

543 

737 

9.91  94 

10' 

34° 

9.7  476 

9.8  290  0.08  14 

0.2  524 

0.1  710 

9.91  86 

56° 

10' 

494 

317     23 

506 

683 

77 

50' 

20' 

513 

344     31 

487 

656 

69 

40' 

30' 

531 

371     40 

469 

629 

60 

30' 

40' 

550 

398     49 

450 

602 

51 

20' 

50' 

568 

425     58 

432 

575 

42 

10' 

35° 

9.7  586 

9.8  452  0.08  66 

0.2  414 

0.1  548 

9.91  34 

55° 

L  cos/ 

Lcoty    log  CSC/ 

log  sec/ 

log  tan/ 

L  sin/ 

/ 

LOGARITHMS   OF   FUNCTIONS. 


XV 


X 

Lsinjr 

/.tanr 

log  sec  X 

log  CSC  X 

log  cot  X 

/.cosjr 

35° 

9.7  586 

9.8  452 

0.08  66 

0.2  414 

0.1  548 

9.91  34 

55° 

10' 

604 

479 

75 

396 

521 

25 

50' 

20' 

622 

506 

84 

378 

494 

16 

40' 

30' 

640 

533 

93 

360 

467 

07 

30' 

40' 

657 

559 

0.09  02 

343 

441 

9.90  98 

20' 

50' 

675 

586 

11 

325 

414 

89 

10' 

36° 

9.7  692 

9.8  613 

0.09  20 

0.2  308 

0.1387 

9.90  80 

54° 

10' 

710 

639 

30 

290 

361 

70 

50' 

ZO' 

727 

666 

39 

273 

334 

61 

40' 

30' 

744 

692 

48 

256 

308 

52 

30' 

40' 

761 

718 

58 

239 

282 

42 

20' 

50' 

778 

745 

67 

222 

255 

33 

10' 

37° 

9.7  795 

9.8  771 

0.09  77 

0.2  205 

0.1  229 

9.90  23 

53° 

10' 

811 

797 

86 

189 

203 

14 

50' 

20' 

828 

824 

96 

172 

176 

04 

40' 

30' 

844 

850 

0.10  05 

156 

150 

9.89  95 

30' 

40' 

861 

876 

15 

139 

124 

85 

20' 

50' 

877 

902 

25 

123 

098 

75 

10' 

38^ 

9.7  893 

9.8  928 

0.10  35 

0.2  107 

0.1  072 

9.89  65 

52° 

10' 

910 

954 

45 

090 

046 

55 

50' 

20' 

926 

•980 

55 

074 

020 

45 

40' 

30' 

941 

9.9  006 

65 

059 

0.0  994 

35 

30' 

40' 

957 

032 

75 

043 

968 

25 

20' 

50' 

973 

058 

85 

027 

942 

15 

10' 

39° 

9.7  989 

9.9  084 

0.10  95 

0.2  011 

0.0  916 

9.89  05 

51° 

10' 

9.8  004 

110 

0.11  05 

0.1  996 

890 

9.88  95 

50' 

20' 

020 

135 

16 

980 

865 

84 

40' 

30' 

035 

161 

26 

965 

839 

74 

30' 

40' 

050 

187 

36 

950 

813 

64 

20' 

50' 

066 

212 

47 

934 

788 

53 

10' 

40° 

9.8  081 

9.9  238 

0.11  57 

0.1  919 

0.0  762 

9.88  43 

50° 

10' 

096 

264 

68 

904 

736 

32 

50' 

20' 

111 

289 

79 

889 

711 

21 

40' 

30' 

125 

315 

90 

875 

685 

10 

30' 

40' 

140 

341 

0.12  00 

860 

659 

00 

20' 

50' 

155 

366 

11 

845 

634 

9.87  89 

10' 

41° 

9.8  169 

9.9  392 

0.12  22 

0.1  831 

0.0  608 

9.87  78 

49° 

10' 

184 

417 

ZZ 

816 

583 

67 

50' 

20' 

198 

443 

44 

802 

557 

56 

40' 

30' 

213 

468 

55 

787 

532 

45 

30' 

40' 

227 

494 

67 

773 

506 

33 

20' 

50' 

241 

519 

78 

759 

481 

22 

10' 

42° 

9.8  255 

9.9  544 

0.12  89 

0.1  745 

0.0  456 

9.87  11 

48° 

Z.COS/ 

/.cot/ 

log  CSC/ 

log  sec/  logtany 

L  sin/ 

/ 

XVI 


LOGARITHMS  OF  FUNCTIONS. 


X 

Lsinx 

Lt&nx 

log  sec  X 

log  CSC  X 

log  cot  ;f 

L  cosx 

48° 

42° 

9.8  255 

9.9  544 

0.12  89 

0.1  745 

0.0  456 

9.87  11 

10' 

269 

570 

0.13  01 

731 

430 

9.86  99 

50' 

20' 

283 

595 

12 

717 

405 

88 

40' 

30' 

297 

621 

24 

703 

379 

76 

30' 

40' 

311 

646 

35 

689 

354 

65 

20' 

50' 

324 

671 

47 

676 

329 

53 

10' 

43° 

9.8  338 

9.9  697 

0.13  59 

0.1  662 

0.0  303 

9.86  41 

47° 

10' 

351 

722 

71 

649 

278 

29 

50' 

20' 

365 

747 

82 

635 

253 

18 

40' 

30' 

378 

772 

94 

622 

228 

06 

30' 

40' 

391 

798 

0.14  06 

609 

202 

9.85  94 

20' 

50' 

405 

823 

18 

595 

177 

82 

10' 

44° 

9.8  418 

9.9  848 

0.14  31 

0.1  582 

0.0  152 

9.85  69 

46° 

10' 

431 

874 

43 

569 

126 

57 

50' 

20' 

444 

899 

55 

556 

101 

45 

40' 

30' 

457 

9.9  924 

68 

543 

0.0  076 

32 

30' 

40' 

469 

949 

80 

531 

051 

20 

20' 

50' 

482 

975 

93 

518 

025 

07 

10' 

45° 

9.8  495 

10.0  000 

0.15  05 

0.1  505 

0.0  000 

9.84  95 

45° 

Z.COS/ 

L  cot/ 

log  CSC/ 

log  sec/  log  tan/ 

/.sin/ 

/ 

FORMUL.^. 

a  CSC  J.=  &  CSC  5=0  CSC  C. 
a^=b^-\-c'^-2  be  cos  A. 
CSC  A 


cot  B=c 


•cot  A 


^-■J(s-a)is-b)(s-c) 


cotlA= ;   cotlB= 

2  >  2 


Circle:  circuni  =  2  7rr;  area^Trr^. 
sector  =:^rx  arc;  seg^ir^^— sin^). 
Sphere:  sur=4  7r^2.  vol=f  wB^. 
zone=2  IT Bh;  seg=l  wJi^ (3  B-h). 
Natural  log  iV=2.30259  logio  K 
Pendulum:  ■nH=gt. 
Falling  bodies:  2s=gt'^=vt=v^-T-g. 


Constants. 

NUMBER. 

LOG. 

radian 

57.29578° 

1.75812 

TT 

180° 

2.25527 

TT 

10800' 

4.03342 

IT 

648000" 

5.81158 

TT 

3.14159 

0.49715 

l-TT 

0.31831 

9.50285 

7r2 

9.86960 

0.99430 

V^r 

1.77245 

0.24857 

1  mile  5280  ft.  3.72263 

1  acre  43560  sq.  ft.  4.63909 

e  2.71828  0.43429 

l^M  2.30259  0.36221 

vel.  light  186330  5.27028 

K40°)  39.0986  1.59216 

^(40°)  32.1573  ~        1.50728 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.00  ON  THE  SEVENTH  DAY 
OVERDUE. 


DEC 


4aE4m 


IML 


^88-f 


X^W^'^^ 


w 


2Nu'&UfiH 


^i:tQl    ^1 


iOctsau 


— eJ^pi^52IllL 


23Mar5  2L . 


2lNov57P" 


REC'D  >.d: 


NOV  15  gs7 


LD  21-100?n-12,'43  (8796s) 


' D     I / I Ou 


aA.53\ 

W5'4 


THE  UNIVERSITY  OF  CAUFORNIA  UBRARY 


